How can I integrate $ int frac{x^2}{1 + x^3} $?












1















$$int frac{x^2}{1 + x^3}$$




I found this problem in one of my past papers and didn't understand how the solution to this is $ frac{1}{3} ln (1+x^3) $. I would really appreciate it if someone guided me through this. I'm sure this is the answer of the integration because I checked it on WolframAlpha and in the mark scheme, but did not understand how to arrive at the answer. Thanks in advance!






calculus integration







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  • Did you try to make a substitution?
    – T. Bongers
    Jan 23 '16 at 7:02










  • Haven't tried it yet. Thanks, I'm on it!
    – Krunal Rindani
    Jan 23 '16 at 7:03






  • 1




    $$dfrac{d(1+x^3)}{dx}=?$$
    – lab bhattacharjee
    Jan 23 '16 at 7:03










  • @T. Bongers I'm not getting it even after substitution.
    – Krunal Rindani
    Jan 23 '16 at 7:14










  • @labbhattacharjee didn't get you
    – Krunal Rindani
    Jan 23 '16 at 7:14
















1















$$int frac{x^2}{1 + x^3}$$




I found this problem in one of my past papers and didn't understand how the solution to this is $ frac{1}{3} ln (1+x^3) $. I would really appreciate it if someone guided me through this. I'm sure this is the answer of the integration because I checked it on WolframAlpha and in the mark scheme, but did not understand how to arrive at the answer. Thanks in advance!






calculus integration







share|cite|improve this question
























  • Did you try to make a substitution?
    – T. Bongers
    Jan 23 '16 at 7:02










  • Haven't tried it yet. Thanks, I'm on it!
    – Krunal Rindani
    Jan 23 '16 at 7:03






  • 1




    $$dfrac{d(1+x^3)}{dx}=?$$
    – lab bhattacharjee
    Jan 23 '16 at 7:03










  • @T. Bongers I'm not getting it even after substitution.
    – Krunal Rindani
    Jan 23 '16 at 7:14










  • @labbhattacharjee didn't get you
    – Krunal Rindani
    Jan 23 '16 at 7:14














1












1








1


1






$$int frac{x^2}{1 + x^3}$$




I found this problem in one of my past papers and didn't understand how the solution to this is $ frac{1}{3} ln (1+x^3) $. I would really appreciate it if someone guided me through this. I'm sure this is the answer of the integration because I checked it on WolframAlpha and in the mark scheme, but did not understand how to arrive at the answer. Thanks in advance!






calculus integration







share|cite|improve this question
















$$int frac{x^2}{1 + x^3}$$




I found this problem in one of my past papers and didn't understand how the solution to this is $ frac{1}{3} ln (1+x^3) $. I would really appreciate it if someone guided me through this. I'm sure this is the answer of the integration because I checked it on WolframAlpha and in the mark scheme, but did not understand how to arrive at the answer. Thanks in advance!






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 '16 at 7:03









SchrodingersCat

22.3k52862




22.3k52862










asked Jan 23 '16 at 7:01









Krunal RindaniKrunal Rindani

143110




143110












  • Did you try to make a substitution?
    – T. Bongers
    Jan 23 '16 at 7:02










  • Haven't tried it yet. Thanks, I'm on it!
    – Krunal Rindani
    Jan 23 '16 at 7:03






  • 1




    $$dfrac{d(1+x^3)}{dx}=?$$
    – lab bhattacharjee
    Jan 23 '16 at 7:03










  • @T. Bongers I'm not getting it even after substitution.
    – Krunal Rindani
    Jan 23 '16 at 7:14










  • @labbhattacharjee didn't get you
    – Krunal Rindani
    Jan 23 '16 at 7:14


















  • Did you try to make a substitution?
    – T. Bongers
    Jan 23 '16 at 7:02










  • Haven't tried it yet. Thanks, I'm on it!
    – Krunal Rindani
    Jan 23 '16 at 7:03






  • 1




    $$dfrac{d(1+x^3)}{dx}=?$$
    – lab bhattacharjee
    Jan 23 '16 at 7:03










  • @T. Bongers I'm not getting it even after substitution.
    – Krunal Rindani
    Jan 23 '16 at 7:14










  • @labbhattacharjee didn't get you
    – Krunal Rindani
    Jan 23 '16 at 7:14
















Did you try to make a substitution?
– T. Bongers
Jan 23 '16 at 7:02




Did you try to make a substitution?
– T. Bongers
Jan 23 '16 at 7:02












Haven't tried it yet. Thanks, I'm on it!
– Krunal Rindani
Jan 23 '16 at 7:03




Haven't tried it yet. Thanks, I'm on it!
– Krunal Rindani
Jan 23 '16 at 7:03




1




1




$$dfrac{d(1+x^3)}{dx}=?$$
– lab bhattacharjee
Jan 23 '16 at 7:03




$$dfrac{d(1+x^3)}{dx}=?$$
– lab bhattacharjee
Jan 23 '16 at 7:03












@T. Bongers I'm not getting it even after substitution.
– Krunal Rindani
Jan 23 '16 at 7:14




@T. Bongers I'm not getting it even after substitution.
– Krunal Rindani
Jan 23 '16 at 7:14












@labbhattacharjee didn't get you
– Krunal Rindani
Jan 23 '16 at 7:14




@labbhattacharjee didn't get you
– Krunal Rindani
Jan 23 '16 at 7:14










2 Answers
2






active

oldest

votes


















3














Hint:



$$u=x^3+1implies du=3x^2,dx$$






share|cite|improve this answer





















  • Got it. Thanks! I messed up while taking $ du $. Thanks!
    – Krunal Rindani
    Jan 23 '16 at 7:18



















0














The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.



Here you have a rational function of the form:



begin{equation}
f(x) = frac{P_n(x)}{Q_m(x)}
end{equation}



Were $P_n(x)$ and $Q_m(x)$ are real valued polynomials of orders $n$ and $m$ respectively. The first step is to factor $Q_m(x)$. By the fundamental theorem of algebra we can factor $Q_m(x)$ is a product of constant, linear, and quadratic terms. For your question:



begin{equation}
f(x) = frac{x^2}{x^3 + 1} rightarrow P_n(x) = P_2(x) = x^2,quad Q_m(x) = Q_3(x) = x^3 + 1
end{equation}



Here $Q_3(x)$ is factored quite easily:



begin{equation}
Q_m(x) = Q_3(x) = x^3 + 1 = left(x + 1right)left(x^2 - x + 1right)
end{equation}



Thus,



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)}
end{equation}



We now apply a Partial Fraction Decomposition to yield:



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)} = frac{1}{3}cdot frac{1}{x + 1} - frac{1}{3}cdot frac{2x - 1}{x^2 - x + 1}
end{equation}



And hence,



begin{align}
int frac{x^2}{x^3 + 1}:dx &= frac{1}{3}int frac{1}{x + 1}:dx + frac{1}{3}int frac{2x - 1}{x^2 - x + 1}:dx \
&= frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right| + C
end{align}



Where $C$ is the constant of integration.



As before, this is overly complicated for this exact situation, but it is a process that is useful in tackling these types of integrals.



Edit I had an incorrect partial fraction decomposition which was picked by a commenter. Thank you to that person.






share|cite|improve this answer



















  • 1




    That can't be right, can it? Didn't you notice that your answer is not equal to $frac13ln|1+x^3|+$ constant? (Your partial fraction decomposition is wrong.)
    – TonyK
    Jan 6 at 11:34












  • @TonyK - It is indeed. Thank you for the pickup. I will correct.
    – DavidG
    Jan 6 at 11:40






  • 1




    I think it should be $frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right|=frac13ln|x^3+1|$
    – TheSimpliFire
    Jan 6 at 11:48










  • Yes, thank you @TheSimpliFire. Is now corrected.
    – DavidG
    Jan 6 at 11:49











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














Hint:



$$u=x^3+1implies du=3x^2,dx$$






share|cite|improve this answer





















  • Got it. Thanks! I messed up while taking $ du $. Thanks!
    – Krunal Rindani
    Jan 23 '16 at 7:18
















3














Hint:



$$u=x^3+1implies du=3x^2,dx$$






share|cite|improve this answer





















  • Got it. Thanks! I messed up while taking $ du $. Thanks!
    – Krunal Rindani
    Jan 23 '16 at 7:18














3












3








3






Hint:



$$u=x^3+1implies du=3x^2,dx$$






share|cite|improve this answer












Hint:



$$u=x^3+1implies du=3x^2,dx$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 23 '16 at 7:06









Yagna PatelYagna Patel

6,42912345




6,42912345












  • Got it. Thanks! I messed up while taking $ du $. Thanks!
    – Krunal Rindani
    Jan 23 '16 at 7:18


















  • Got it. Thanks! I messed up while taking $ du $. Thanks!
    – Krunal Rindani
    Jan 23 '16 at 7:18
















Got it. Thanks! I messed up while taking $ du $. Thanks!
– Krunal Rindani
Jan 23 '16 at 7:18




Got it. Thanks! I messed up while taking $ du $. Thanks!
– Krunal Rindani
Jan 23 '16 at 7:18











0














The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.



Here you have a rational function of the form:



begin{equation}
f(x) = frac{P_n(x)}{Q_m(x)}
end{equation}



Were $P_n(x)$ and $Q_m(x)$ are real valued polynomials of orders $n$ and $m$ respectively. The first step is to factor $Q_m(x)$. By the fundamental theorem of algebra we can factor $Q_m(x)$ is a product of constant, linear, and quadratic terms. For your question:



begin{equation}
f(x) = frac{x^2}{x^3 + 1} rightarrow P_n(x) = P_2(x) = x^2,quad Q_m(x) = Q_3(x) = x^3 + 1
end{equation}



Here $Q_3(x)$ is factored quite easily:



begin{equation}
Q_m(x) = Q_3(x) = x^3 + 1 = left(x + 1right)left(x^2 - x + 1right)
end{equation}



Thus,



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)}
end{equation}



We now apply a Partial Fraction Decomposition to yield:



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)} = frac{1}{3}cdot frac{1}{x + 1} - frac{1}{3}cdot frac{2x - 1}{x^2 - x + 1}
end{equation}



And hence,



begin{align}
int frac{x^2}{x^3 + 1}:dx &= frac{1}{3}int frac{1}{x + 1}:dx + frac{1}{3}int frac{2x - 1}{x^2 - x + 1}:dx \
&= frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right| + C
end{align}



Where $C$ is the constant of integration.



As before, this is overly complicated for this exact situation, but it is a process that is useful in tackling these types of integrals.



Edit I had an incorrect partial fraction decomposition which was picked by a commenter. Thank you to that person.






share|cite|improve this answer



















  • 1




    That can't be right, can it? Didn't you notice that your answer is not equal to $frac13ln|1+x^3|+$ constant? (Your partial fraction decomposition is wrong.)
    – TonyK
    Jan 6 at 11:34












  • @TonyK - It is indeed. Thank you for the pickup. I will correct.
    – DavidG
    Jan 6 at 11:40






  • 1




    I think it should be $frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right|=frac13ln|x^3+1|$
    – TheSimpliFire
    Jan 6 at 11:48










  • Yes, thank you @TheSimpliFire. Is now corrected.
    – DavidG
    Jan 6 at 11:49
















0














The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.



Here you have a rational function of the form:



begin{equation}
f(x) = frac{P_n(x)}{Q_m(x)}
end{equation}



Were $P_n(x)$ and $Q_m(x)$ are real valued polynomials of orders $n$ and $m$ respectively. The first step is to factor $Q_m(x)$. By the fundamental theorem of algebra we can factor $Q_m(x)$ is a product of constant, linear, and quadratic terms. For your question:



begin{equation}
f(x) = frac{x^2}{x^3 + 1} rightarrow P_n(x) = P_2(x) = x^2,quad Q_m(x) = Q_3(x) = x^3 + 1
end{equation}



Here $Q_3(x)$ is factored quite easily:



begin{equation}
Q_m(x) = Q_3(x) = x^3 + 1 = left(x + 1right)left(x^2 - x + 1right)
end{equation}



Thus,



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)}
end{equation}



We now apply a Partial Fraction Decomposition to yield:



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)} = frac{1}{3}cdot frac{1}{x + 1} - frac{1}{3}cdot frac{2x - 1}{x^2 - x + 1}
end{equation}



And hence,



begin{align}
int frac{x^2}{x^3 + 1}:dx &= frac{1}{3}int frac{1}{x + 1}:dx + frac{1}{3}int frac{2x - 1}{x^2 - x + 1}:dx \
&= frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right| + C
end{align}



Where $C$ is the constant of integration.



As before, this is overly complicated for this exact situation, but it is a process that is useful in tackling these types of integrals.



Edit I had an incorrect partial fraction decomposition which was picked by a commenter. Thank you to that person.






share|cite|improve this answer



















  • 1




    That can't be right, can it? Didn't you notice that your answer is not equal to $frac13ln|1+x^3|+$ constant? (Your partial fraction decomposition is wrong.)
    – TonyK
    Jan 6 at 11:34












  • @TonyK - It is indeed. Thank you for the pickup. I will correct.
    – DavidG
    Jan 6 at 11:40






  • 1




    I think it should be $frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right|=frac13ln|x^3+1|$
    – TheSimpliFire
    Jan 6 at 11:48










  • Yes, thank you @TheSimpliFire. Is now corrected.
    – DavidG
    Jan 6 at 11:49














0












0








0






The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.



Here you have a rational function of the form:



begin{equation}
f(x) = frac{P_n(x)}{Q_m(x)}
end{equation}



Were $P_n(x)$ and $Q_m(x)$ are real valued polynomials of orders $n$ and $m$ respectively. The first step is to factor $Q_m(x)$. By the fundamental theorem of algebra we can factor $Q_m(x)$ is a product of constant, linear, and quadratic terms. For your question:



begin{equation}
f(x) = frac{x^2}{x^3 + 1} rightarrow P_n(x) = P_2(x) = x^2,quad Q_m(x) = Q_3(x) = x^3 + 1
end{equation}



Here $Q_3(x)$ is factored quite easily:



begin{equation}
Q_m(x) = Q_3(x) = x^3 + 1 = left(x + 1right)left(x^2 - x + 1right)
end{equation}



Thus,



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)}
end{equation}



We now apply a Partial Fraction Decomposition to yield:



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)} = frac{1}{3}cdot frac{1}{x + 1} - frac{1}{3}cdot frac{2x - 1}{x^2 - x + 1}
end{equation}



And hence,



begin{align}
int frac{x^2}{x^3 + 1}:dx &= frac{1}{3}int frac{1}{x + 1}:dx + frac{1}{3}int frac{2x - 1}{x^2 - x + 1}:dx \
&= frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right| + C
end{align}



Where $C$ is the constant of integration.



As before, this is overly complicated for this exact situation, but it is a process that is useful in tackling these types of integrals.



Edit I had an incorrect partial fraction decomposition which was picked by a commenter. Thank you to that person.






share|cite|improve this answer














The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.



Here you have a rational function of the form:



begin{equation}
f(x) = frac{P_n(x)}{Q_m(x)}
end{equation}



Were $P_n(x)$ and $Q_m(x)$ are real valued polynomials of orders $n$ and $m$ respectively. The first step is to factor $Q_m(x)$. By the fundamental theorem of algebra we can factor $Q_m(x)$ is a product of constant, linear, and quadratic terms. For your question:



begin{equation}
f(x) = frac{x^2}{x^3 + 1} rightarrow P_n(x) = P_2(x) = x^2,quad Q_m(x) = Q_3(x) = x^3 + 1
end{equation}



Here $Q_3(x)$ is factored quite easily:



begin{equation}
Q_m(x) = Q_3(x) = x^3 + 1 = left(x + 1right)left(x^2 - x + 1right)
end{equation}



Thus,



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)}
end{equation}



We now apply a Partial Fraction Decomposition to yield:



begin{equation}
f(x) = frac{x^2}{left(x + 1right)left(x^2 - x + 1right)} = frac{1}{3}cdot frac{1}{x + 1} - frac{1}{3}cdot frac{2x - 1}{x^2 - x + 1}
end{equation}



And hence,



begin{align}
int frac{x^2}{x^3 + 1}:dx &= frac{1}{3}int frac{1}{x + 1}:dx + frac{1}{3}int frac{2x - 1}{x^2 - x + 1}:dx \
&= frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right| + C
end{align}



Where $C$ is the constant of integration.



As before, this is overly complicated for this exact situation, but it is a process that is useful in tackling these types of integrals.



Edit I had an incorrect partial fraction decomposition which was picked by a commenter. Thank you to that person.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 6 at 11:49

























answered Jan 6 at 11:16









DavidGDavidG

1,930620




1,930620








  • 1




    That can't be right, can it? Didn't you notice that your answer is not equal to $frac13ln|1+x^3|+$ constant? (Your partial fraction decomposition is wrong.)
    – TonyK
    Jan 6 at 11:34












  • @TonyK - It is indeed. Thank you for the pickup. I will correct.
    – DavidG
    Jan 6 at 11:40






  • 1




    I think it should be $frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right|=frac13ln|x^3+1|$
    – TheSimpliFire
    Jan 6 at 11:48










  • Yes, thank you @TheSimpliFire. Is now corrected.
    – DavidG
    Jan 6 at 11:49














  • 1




    That can't be right, can it? Didn't you notice that your answer is not equal to $frac13ln|1+x^3|+$ constant? (Your partial fraction decomposition is wrong.)
    – TonyK
    Jan 6 at 11:34












  • @TonyK - It is indeed. Thank you for the pickup. I will correct.
    – DavidG
    Jan 6 at 11:40






  • 1




    I think it should be $frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right|=frac13ln|x^3+1|$
    – TheSimpliFire
    Jan 6 at 11:48










  • Yes, thank you @TheSimpliFire. Is now corrected.
    – DavidG
    Jan 6 at 11:49








1




1




That can't be right, can it? Didn't you notice that your answer is not equal to $frac13ln|1+x^3|+$ constant? (Your partial fraction decomposition is wrong.)
– TonyK
Jan 6 at 11:34






That can't be right, can it? Didn't you notice that your answer is not equal to $frac13ln|1+x^3|+$ constant? (Your partial fraction decomposition is wrong.)
– TonyK
Jan 6 at 11:34














@TonyK - It is indeed. Thank you for the pickup. I will correct.
– DavidG
Jan 6 at 11:40




@TonyK - It is indeed. Thank you for the pickup. I will correct.
– DavidG
Jan 6 at 11:40




1




1




I think it should be $frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right|=frac13ln|x^3+1|$
– TheSimpliFire
Jan 6 at 11:48




I think it should be $frac{1}{3}lnleft|x + 1right| + frac{1}{3} cdot lnleft|x^2 - x + 1 right|=frac13ln|x^3+1|$
– TheSimpliFire
Jan 6 at 11:48












Yes, thank you @TheSimpliFire. Is now corrected.
– DavidG
Jan 6 at 11:49




Yes, thank you @TheSimpliFire. Is now corrected.
– DavidG
Jan 6 at 11:49


















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