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I tried solving the following integral using integral by parts : $$\int \frac{x^3}{(x^2+1)^2}dx$$ but I got a different answer from Wolfram Calculator This is the answer that I got : $$\int \frac{x^3}{(x^2+1)^2}dx=\frac{1}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C$$ I am wondering if I am wrong or the calculator if I am where did I do something wrong.

Here is my solution : $$\int \frac{x^3}{(x^2+1)^2}dx$$ $$\int x^2\frac{x}{(x^2+1)^2}dx$$ $\implies u =x^2 \qquad u'=2x$
$\displaystyle\implies v'=\frac{x}{(x^2+1)^2}\qquad v=\int\frac{x}{(x^2+1)^2}dx$ $$\int\frac{x}{(x^2+1)^2}dx$$ $\implies \zeta=x^2+1$
$\implies\displaystyle \frac{d\zeta}{2}=x$ $$\boxed{v=\frac{1}{2}\int\frac{d\zeta}{\zeta^2}=-0.5\zeta^{-1}=\frac{-1}{2(x^2+1)}}$$ $$\int x^2\frac{x}{(x^2+1)^2}dx=x^2\frac{-1}{2(x^2+1)}-\underbrace{(\int \frac{-2x}{2(x^2+1)}dx)}_{-\frac{1}{2}\ln(x^2+1)}$$ $$\boxed{\int \frac{x^3}{(x^2+1)^2}dx=\frac{-x^2}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C}$$

zlmk
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2 Answers2

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Both are correct. $\frac{-x^2}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C=\frac 1{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C'$ where $C'=C-\frac 1 2$

[Just transfer $\frac 1{2(x^2+1)}$ to the left side and combine it with $\frac{-x^2}{2(x^2+1)}$].

Kavi Rama Murthy
  • 359,332
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BOTH are correct since the two answers differ by a constant.

$$ \begin{aligned} \int \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x &=-\frac{1}{2} \int x^{2} d\left(\frac{1}{x^{2}+1}\right) \\ &=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x \\ &=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\ &=-\frac{x^{2}+1-1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\ &=-\frac{1}{2}+\frac{1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\ &=\frac{1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{2}, \end{aligned} $$

where $C_{2}= C_{1} -\frac{1}{2}$.

Lai
  • 31,615