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

Question

$$\int \frac{1}{(x^2+1)^2}\mathrm dx$$ Look simple, but I stuck a little bit. What method is better in this situation? Please, help.

Answer 1

Hint: $$x = \tan\theta\implies dx = \sec^2 \theta\ d\theta$$ $$\int\frac{dx}{\left(1+x^2\right)^2} = \int\frac{\sec^2\theta\ d\theta}{\left(1+\tan^2\theta\right)^2} = \dots$$

Answer 2

There is a faster way. Substitute

$$x = \tan(z) ~~~~~~~ \text{d}x = \sec^2(z)\ \text{d}z$$

thus

$$(x^2+1)^2 = (\tan^2(z) + 1)^2 = \sec^4(z) ~~~~~~~~~~~ z = \arctan(x)$$

And remembering that $$\frac{1}{\sec^2} = \cos^2$$ your integral is simply

$$\int\cos^2(z)\ \text{d}z$$

Which is trivial and left to you.

The tangent/secant substitution is a great technique which most of people ignore. Study it, and you will solve lots of awesome integrals!

Final Result

$$\frac{x^2\arctan(x) + x + \arctan(x)}{2x^2 + 2}$$

Answer 3

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

Set $x:=\tan(u)$ and $dx=\sec^2(u)du$ . then $(x^2+1)^2=(\tan^2(u)+1)^2=\sec^4(u)$ and $u=\arctan(x)$

$$=\int \cos^2(u)du=\frac 1 2 \int\cos(2u)du+\frac 1 2\int 1 du=\frac u 2+\frac 1 4\sin(2u)+C=\boxed{\frac{x^2\arctan(x)+x+\arctan(x)}{2x^2+2}+C}$$

Answer 4

Let us find $\dfrac{d\{(ax+b)/(x^2+1)^n\}}{dx}$

$$=\dfrac{-ax^2-2bx+a}{(x^2+1)^{n+1}}=\dfrac{-a(x^2+1)-2bx+2a}{(x^2+1)^{n+1}}$$

$$=-\dfrac a{(x^2+1)^n}+\dfrac{2a-2bx}{(x^2+1)^{n+1}}$$

Integrating both sides, $$b=0\implies2\int\dfrac{dx}{(x^2+1)^{n+1}}=\int\dfrac{dx}{(x^2+1)^n}+\dfrac x{(x^2+1)^n}$$

$$n=1\implies2\int\dfrac{dx}{(x^2+1)^2}=\int\dfrac{dx}{x^2+1}+\dfrac x{(x^2+1)}$$