Julia White

2022-01-03

How to calculate

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

Terry Ray

Beginner2022-01-04Added 50 answers

First, you used parts: with $u=x$ and $dv=\frac{x}{{({x}^{2}+1)}^{2}}dx$ , you got

$\int \frac{{x}^{2}}{{({x}^{2}+1)}^{2}}dx=xv-\int vdx$ (1)

From here, I suggested the substitution$t={x}^{2}$ in order to integrate dv and get v. Doing this, we see that $dt=2xdx$ , so $xdx=\frac{1}{2}dt$ . Thus

$v=\int dv=\int \frac{x}{{({x}^{2}+1)}^{2}}dx=\frac{1}{2}\int \frac{dt}{{(t+1)}^{2}}$

This latter integral is easy: if you can't guess it, let$s=t+1$ . In any case, we get

$v=-\frac{1}{2(t+1)}=-\frac{1}{2({x}^{2}+1)}$

Going back to (1),

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

and you should recognize the final integral as$\mathrm{arctan}x$ . Putting it all together, the answer is

$\frac{1}{2}\mathrm{arctan}x-\frac{x}{2({x}^{2}+1)}+C$

From here, I suggested the substitution

This latter integral is easy: if you can't guess it, let

Going back to (1),

and you should recognize the final integral as

Daniel Cormack

Beginner2022-01-05Added 34 answers

HINT:

Use trigonometric substitution$x=\mathrm{tan}\theta$

$\int \frac{{x}^{2}}{{({x}^{2}+1)}^{2}}dx=\frac{1}{2}\int (1-\mathrm{cos}2\theta )d\theta$

$\mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta}{1+{\mathrm{tan}}^{2}\theta}$

Use trigonometric substitution

Vasquez

Expert2022-01-09Added 669 answers

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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