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

Would anyone help me calculate the following integral? $$\int \frac{dx}{(x^2+1)^3}$$

During our lecutre we've done very similiar one, $$\int \frac{dx}{(x^2+1)^2}$$ like that:

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

$$= \Biggr\rvert $$\begin{split} & u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\ & u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1} \end{split}$$ \Biggr\rvert$$

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

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

Thank you.

We will find a general reduction formula for the integral $$I_n=\int\frac{dx}{(ax^2+b)^n}$$ Integration by parts with $$dv=dx\Rightarrow v=x\\ u=\frac1{(ax^2+b)^n}\Rightarrow du=\frac{-2anx}{(ax^2+b)^{n+1}}dx$$ Yields $$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2}{(ax^2+b)^{n+1}}dx$$ $$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bn\int\frac{dx}{(ax^2+b)^{n+1}}$$ $$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$ $$I_n=\frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$ $$2bnI_{n+1}=\frac{x}{(ax^2+b)^n}+(2n-1)I_n$$ $$I_{n+1}=\frac{x}{2bn(ax^2+b)^n}+\frac{2n-1}{2bn}I_n$$ replacing $$n+1$$ with $$n$$, $$I_{n}=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}$$ Now for the base case $$I_1$$: $$I_1=\int\frac{dx}{ax^2+b}$$ Let $$x=\sqrt{\frac{b}a}\tan u\Rightarrow dx=\sqrt{\frac{b}a}\sec^2u\, du$$ So $$I_1=\sqrt{\frac{b}a}\int\frac{\sec^2u}{b\tan^2u+b}du$$ $$I_1=\frac1{\sqrt{ab}}\int\frac{\sec^2u}{\sec^2u}du$$ $$I_1=\frac1{\sqrt{ab}}\int du$$ $$I_1=\frac{u}{\sqrt{ab}}$$ $$I_1=\frac1{\sqrt{ab}}\arctan\sqrt{\frac{a}{b}}x+C$$ Plug in your specific $$a,b$$ and $$n$$, and you're good to go.

Edit: Whatever I'll just give you the answer.

Plugging in $$a=1,\ b=1,\ n=3$$ we have $$I_{3}=\frac{x}{4(x^2+1)^{2}}+\frac{3}{4}I_{2}$$ Note that $$I_{2}=\frac{x}{2(x^2+1)}+\frac{1}{2}I_{1}$$ $$I_{2}=\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan x+C$$ So $$I_{3}=\frac{x}{4(x^2+1)^{2}}+\frac{3x}{8(x^2+1)}+\frac{3}{8}\arctan x+C$$

So why not use the same trick? Since you know from lecture what $$f_2(x) = \int \frac{dx}{\left(x^2+1\right)^2}$$ is, you now have $$\begin{split} f_3(x) &= \int \frac{dx}{\left(x^2+1\right)^3}\\ &= \int \frac{dx}{\left(x^2+1\right)^2} - \int x \cdot \frac{xdx}{\left(x^2+1\right)^3}\\ &= f_2(x) - \text{proceed by parts as before} \end{split}$$

Use $$\int \frac{dx}{(x^2+1)^3}=\frac{x}{4(x^2+1)^2}+\frac{3}{4}\int \frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.

I used:

$$p=x^2+1\to dx =\frac{1}{2\sqrt{p-1}}$$ which leads to:

$$\int{\frac{dx}{(x^2+1)^3}=\frac12 \int{\frac{dp}{p\sqrt{p-1}}}}$$ Then the substitution $$q=\sqrt{p-1}\to dp=2q\space dq$$ Leads to $$\frac12\int\frac{dp}{p\sqrt{p-1}}=\frac12\int\frac{2q}{q(q^2+1)}dq=\int\frac{dq}{q^2+1}$$

Which is resolved simply.