Trouble understanding this hard integral $\int \frac{x^3}{(x^2+1)^2}dx$

Question

Here is the integral:

$$\int \frac{x^3}{(x^2+1)^2}dx$$

Now, I was trying $U = x^2+1 $, then $du = 2xdx$:

$$\int \frac{x^3}{u^2} \frac{du}{2x}$$

$$ \frac{1}{2}\int \frac{x^2}{u^2} du$$

How can I continue this integral? I would like to understand how to solve it only using substitution.

Answer 1

Let $\;\;x^2 +1=u$

So you now have, $$\frac{1}{2}\int \frac{u-1}{u^2}\mathrm{d}u$$

which can be integrated as $$\frac{1}{2}( \ln(u) +\frac{1}{u})$$

that is $$\frac{\ln(x^2+1)}{2} +\frac{1}{2(x^2+1)}+C$$

Where $C$ is an arbitrary constant.

Answer 2

$$\frac{x^3dx}{(x^2+1)^2}$$ $$=\frac{1}{2}\frac{(x^2 + 1) d(x^2+1)}{(x^2+1)^2} -\frac{1}{2} \frac{d(x^2 +1)}{(x^2 + 1)^2}$$

It's just a question to see that $x^2 = (x^2+1) - 1$ (and $d(f(x)) = d(f(x) ± a)$ for a fix number $a$ (like $1$)).

Answer 3

$$\int\frac{x^2}{u^2}du=\int\frac{u-1}{u^2}du=\int(\frac{1}{u}-\frac{1}{u^2})du$$

Hope,now you will try .

Answer 4

Integration by parts $$ \begin{aligned} \int \frac{x^2}{\left(x^2+1\right)^2} d x&=-\frac{1}{2} \int x^2 d\left(\frac{1}{x^2+1}\right) \\ & =-\frac{x^2}{2\left(x^2+1\right)}+\int \frac{x}{x^2+1} d x \\ & =-\frac{x^2}{2\left(x^2+1\right)}+\frac{1}{2} \ln \left(x^2+1\right)+C_1 \\ O R&=\frac{1}{2\left(x^2+1\right)}+\frac{1}{2} \ln \left(x^2+1\right)+C_2 \end{aligned} $$

Answer 5

For a trig-sub approach, notice that a factor of the form $x^2+1$ in the denominator suggests the trig substitution $\color{blue}{\tan(\alpha)} = \color{blue}{x}$, since $\tan^2(\alpha) +1 = \sec^2(\alpha)$. Recalling that $\frac{\mathrm{d}}{\mathrm{d}\alpha}\tan(\alpha) = \sec^2(\alpha)$ we get \begin{align*} \int \frac{x^3}{\left(x^2+1\right)^2}\, \mathrm{d}x &= \int \frac{\tan^3(\alpha)}{\sec^4(\alpha)}\sec^2(\alpha) \, \mathrm{d}\alpha \\ & = \int \left(\cos(\alpha)-\frac{1}{\cos(\alpha)}\right)\left(-\sin(\alpha)\right) \, \mathrm{d}\alpha \\ & \overset{\color{purple}{u = \cos(\alpha)}}{=}\int u - \frac1{u} \, \mathrm{d}u\\ & = \frac{u^2}{2} -\ln(u) + C\\ & =\frac{1}{2(x^2+1)}+\frac12\ln\left(x^2+1\right)+C \end{align*} since $u = \cos(\arctan(x)) =$ $\frac{1}{\sqrt{x^2+1}}$.

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The previous solution can be even further condensed combining both substitutions and directly trying $ u = \frac{1}{\sqrt{x^2+1}}$. This gives $\mathrm{d}x = \left(\frac{\mathrm{d}}{\mathrm{d}u}\frac{\sqrt{1-u^2}}{u}\right)\mathrm{d}u = - \frac{1}{u^2 \sqrt{1-u^2}}\, \mathrm{d}u$, resulting in $$ \int \frac{x^3}{\left(x^2+1\right)^2}\, \mathrm{d}x = \int u - \frac 1u\, \mathrm{d}u =\frac{1}{2(x^2+1)}+\frac12\ln\left(x^2+1\right)+C $$

Answer 6

$$\begin{gathered}\;u={x}^{2}+1&\\\;\dfrac{1}{2}\,\mathrm{d}u=x\,\mathrm{d}x&\end{gathered}$$ $$\int{\dfrac{u-1}{2\,{u}^{2}}}{\;\mathrm{d}u}$$ $$\dfrac{1}{2}\int{\dfrac{1}{u}-\dfrac{1}{{u}^{2}}}{\;\mathrm{d}u}$$ $$\dfrac{1}{2}\left(\int{\dfrac{1}{u}}{\;\mathrm{d}u}-\int{\dfrac{1}{{u}^{2}}}{\;\mathrm{d}u}\right)=$$ $$=\dfrac{\ln\left(\left|u\right|\right)}{2}+\dfrac{1}{2\,u}$$ $$\dfrac{\ln\left({x}^{2}+1\right)}{2}+\dfrac{1}{2\,\left({x}^{2}+1\right)}$$

Answer 7

Another possible approach: \begin{align*} \int\frac{x^{3}}{(x^{2} + 1)^{2}}\mathrm{d}x & = \int\frac{\sinh^{3}(u)}{(\sinh^{2}(u) + 1)^{2}}\mathrm{d}(\sinh(u))\\ & = \int\frac{\sinh^{3}(u)}{\cosh^{3}(u)}\mathrm{d}u\\ & = \int\frac{(\cosh^{2}(u) - 1)}{\cosh^{3}(u)}\mathrm{d}(\cosh(u))\\ & = \int\frac{z^{2} - 1}{z^{3}}\mathrm{d}z\\ & = \ln|z| + \frac{1}{2z^{2}} + C\\ & = \ln|\cosh(u)| + \frac{1}{2\cosh^{2}(u)} + C\\ & = \frac{1}{2}\ln(x^{2} + 1) + \frac{1}{2(1 + x^{2})} + C \end{align*}

Answer 8

Using partial fraction decomposition and $u=x^2+1$ (so $du=2x dx$),

$$ \int \frac{x^3}{(x^2+1)^2}dx = \int \frac{x}{x^2+1}-\frac{x}{(x^2+1)^2} dx = \frac{1}{2}\left(\int \frac{du}{u} - \frac{du}{u^2} \right) = \frac{1}{2}\left(\ln|u|+\frac{1}{u}\right)+C $$

so $\int \frac{x^3}{(x^2+1)^2}dx = \frac{1}{2}\left(\ln(x^2+1)+\frac{1}{x^2+1}\right)+C $.