# Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$

Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$

My Attempt

Using Partial fractions $$\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\frac{1}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)}+\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=I_1+I_2\\ I_1=\bigg[\frac{1}{b^2}.\frac{1}{ab}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0=\bigg[\frac{1}{ab^3}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0$$ $$I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=\frac{b^2-a^2}{b^2}\bigg[\frac{x}{(a^2+b^2x^2)^2}-\int\frac{-2.2x.x}{(a^2+b^2x^2)^3}dx\bigg]$$

How do I evaluate the integral $I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$ ?

• let $x = \frac ab \tan \theta$? – steven gregory May 28 '18 at 14:54
• @stevengregory how does that help ? – ss1729 May 28 '18 at 14:59
• This integral is what residue theorem handles extremely well. Note that the integrand is even. – Szeto May 28 '18 at 15:05

$$I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$$

If $$x = \frac ab \tan \theta$$, then

• $$(a^2+b^2x^2)=a^2(1+tan^2 \theta)=a^2 sec^2 \theta$$

• $$dx = \frac ab sec^2 \theta \ d\theta$$

• $$\displaystyle \int \frac{dx}{(a^2+b^2x^2)^2} = \int \frac{\frac ab sec^2 \theta \ d\theta}{a^4 sec^4 \theta} = \int \frac{1}{a^3b}\cos^2 \theta \ d\theta$$

To evaluate $\int\limits_0^{+\infty}\frac{x^2\,dx}{(a^2+b^2x^2)^2}$ just take $\int\limits_0^{+\infty}\frac{dx}{a^2+b^2x^2}$ by parts.

And $\int\limits_0^{+\infty}\frac{dx}{(a^2+b^2x^2)^2}$ reduces to the previous integral by substitution $t=\frac{a}{bx}$.

$\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}+\int_0^\infty\frac{x^2dx}{(a^2+b^2x^2)^2}=I_1+I_2$

Without bounds and constant of integration we have putting $x=\dfrac {at}{b}$ with integration by parts $$I_1=\frac ab\left(\frac{t}{(a^2+b^2t^2)^2}-\int\frac{-4a^2t}{(a^2+a^2t^2)^3}tdt\right)$$ $$I_1=\frac ab\left(\frac{t}{(a^2+b^2t^2)^2}+\frac{1}{2a^4}(\arctan(t)-\frac 14\sin(4\arctan(t))\right)$$ It follows coming back to $x$, $$I_1=\frac{x}{a^2+b^2x^2)^2}+\frac{1}{2ba^3}\arctan\left(\frac{bx}{a}\right)-\frac{1}{2ba^3}\frac14\sin(4\arctan\left(\frac{bx}{a}\right)$$

$I_2$ is easier and we have

$$I_2=\frac{1}{2ab^3}\arctan\left(\frac{bx}{a}\right)-\frac{1}{4ab^3}\sin\left(2\arctan(\frac{bx}{a}\right)$$

I leave the calculations with the bounds $0$ and $\infty$ to finish.