Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}D\sec^2\theta d\theta
= \frac{1}{D^3}\int_{-\infty}^\infty \cos^2 \theta d\theta$$
$$=\frac{1}{D^3}[\frac{\theta}{2} + \frac{\sin {2\theta}}{4} + C]_{-\infty}^\infty$$
Put $\theta = \arctan{\frac{x}{D}}$;
$$=\frac{1}{D^3}[\frac{\arctan{\frac{x}{D}}}{2} + \frac{\sin {(2\arctan{\frac{x}{D})}}}{4} + C]_{-\infty}^\infty$$
We get the integral as $\frac{\pi}{2D^3}$. Since $\lim_{x \to +\infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = \frac{-\pi}{2}$
I feel something isn't right here. Can anyone point the mistake please. Thank you very much.

I’ll suggest another method, involving Integration By Parts. Let the integral to be found be denoted by J.   We start by performing an integration by parts of the function $\frac{1}{x^2+D^2}$:
$$I= \int \frac{1}{x^2+D^2} dx=  \frac{x}{x^2+D^2}-\int \left(-\frac{x(2x)}{(x^2+D^2)^2}\right)dx$$ $$=  \frac{x}{x^2+D^2}+2 \int\frac{x^2+D^2-D^2}{(x^2+D^2)^2}dx$$$$=  \frac{x}{x^2+D^2}+2(I-D^2J)$$ so that we have finally, $$J=\frac{1}{2D^2}\left( \frac{x}{x^2+D^2}+I\right)$$ $$J=\frac{1}{2D^2}\left( \frac{x}{x^2+D^2}+\frac 1D \arctan\left(\frac xD\right)\right)$$ which, on putting the limits, gives $\displaystyle \frac{\pi}{2D^3}$.
