Can you solve $\int 1/(x^2+1)^2\, dx$

Question

I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$ to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$ so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$ $$1+0x^2=(A+B)x^2+(A+B)1$$ Thus I got this system of equations: $$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.

Answer 1

If we use the substitution $x = \tan \theta$, $dx = \sec^2 \theta \,d\theta$, the integral becomes \begin{align*} \int \frac{1}{(x^2+1)^2}\, dx &= \int \frac{\sec^2 \theta}{(\tan^2\theta + 1)^2}\,d\theta\\ &= \int \frac{\sec^2\theta}{\sec^4 \theta}\,d\theta\\ &= \int\cos^2\theta \,d\theta\\ &= \frac{1}{2}(\theta + \sin\theta\cos\theta) + C. \end{align*} I will leave that last integral to you, but we can use the double angle formula for cosine. So, we just need to undo the substitution. Given that $x = \tan\theta$, we can conclude that $\theta = \arctan x$ and $$\sin\theta = \frac{x}{\sqrt{1+x^2}}\quad\text{and}\quad \cos\theta = \frac{1}{\sqrt{1+x^2}}.$$ This follows from the fact that $$x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta \implies \cos^2\theta = \frac{1}{1+x^2}.$$

This gives us $$\int \frac{1}{(x^2+1)^2}\, dx = \frac{1}{2}(\theta + \sin\theta\cos\theta) +C= \frac{1}{2}\left(\arctan x + \frac{x}{1+x^2}\right) +C.$$

Answer 2

Integrate as follows

\begin{align} \int {\dfrac{1}{(x^2+1)^2}}\,dx &= \int {\dfrac{1}{x^2+1}}\,dx - \int {\dfrac{x^2}{(x^2+1)^2}}\,dx\\ &= \int {\dfrac{1}{x^2+1}}\,dx + \frac1{2}\int x\>d\left({\dfrac{1}{x^2+1}}\right)\\ &= \frac1{2} \frac x{x^2+1} + \frac1{2}\int {\dfrac{1}{x^2+1}}\,dx\\ &= \frac1{2} \frac x{x^2+1} + \frac1{2}\tan^{-1} x+C \end{align}

Answer 3

There's multiple ways to do this integral including IBP, substitution... Here's one that uses Feynman's trick (differentiation under the integral sign)

We know the following antiderivative

$$I(a)=\int^x\frac{ dx'}{x'^2+a^2}=\frac{1}{a}\arctan(x/a)$$

You can see we introduced a parameter that did not exist before. We differentiate under the integral sign to obtain the desired integral form

$$J(a)=\int^x\frac{dx'}{(x'^2+a^2)^2}=-\frac{1}{2a}\frac{\partial I}{\partial a}=\frac{1}{2a^3}\left(\arctan(x/a)+\frac{a x}{x^2+a^2}\right)$$

$J(1)$ is the indefinite integral we seek.

EDIT: Here's a way that utilizes IBP.

Rewrite by adding and subtracting $x^2$ in the numerator

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

However note that

$$\int xdx\frac{x}{(x^2+1)^2}=-\frac{1}{2}\int dx~x \frac{d}{dx}\left(\frac{1}{x^2+1}\right)=-\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan x$$

which yields the desired result

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

Answer 4

You cant simplify with PFD on something like this, you do have a few choices though:

1. Trigonometric substitution
2. manipulate the expression

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I think the Trig is fairly well covered so I will do how to manipulate the expression: $$\frac{1}{(1+x^2)^2}=\frac{1+x^2-x^2}{(1+x^2)^2}=\frac{1+x^2}{(1+x^2)^2}-\frac{x^2}{(1+x^2)^2}$$ $$=\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}=\frac1{1+x^2}-x\frac{x}{(1+x^2)^2}$$ Now the first part of this is a standard integral and the second can be done using integration by parts, as shown well by @Quanto now that I check.

Answer 5

$I=\int \frac{dx}{(x^2+1)^2}=\frac{1}{2} \int \frac{1+x^2+1-x^2}{x^4+2x^2+1}.dx$

$\implies I= \frac{1}{2} \left ( \int \frac{1+x^2}{x^4+2x^2+1}.dx +\int \frac{1-x^2}{x^4+2x^2+1}.dx\right).$

$I_1=\int \frac{1+x^2}{x^4+2x^2+1}.dx=\int \frac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}.dx=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+4}.dx$

Now $t=x-\frac 1 x \implies dt=\left(1+\frac{1}{x^2} \right)dx$

$I_1=\int\frac{dt}{t^2+4}=\frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right)$.

$I_2=\int \frac{1-x^2}{x^4+2x^2+1}.dx=-\int \frac{1-\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}.dx=-\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2+1}.dx$

Now $p=x+\frac 1 x \implies dp=\left(1-\frac{1}{x^2} \right)dx$

$I_2=\int\frac{dp}{p^2+1}=\tan^{-1}(p)$

$I=I_1+I_2=\frac{1}{2}\tan^{-1}\left(\frac{x-\frac 1 x}{2}\right)+\tan^{-1}(x+\frac 1 x).$