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Given $\int^x_0 \frac{dx}{x^2 + a^2} = \frac{1}{a} tan^-1 (\frac{x}{a})$, using DUIS find the value of $\int^x_0 \frac{dx}{(x^2 + a^2)^2}$
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Solution:

$\int_{0}^{x} \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)$

Differentiate w.r.t a, taking 'a' as parameter

$\frac{d}{d a} \int_{0}^{x} \frac{1}{x^{2}+a^{2}} d x=\frac{d}{d a}\left[\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]$

Applying D.U.L.S rule,

D.U.1.S rule says that if function and its partial derivative is continuous then we can apply differential operator in the integral operator by converting it into partial derivative taking one parameter fro function.

$\int_{0}^{x} \frac{\partial}{\partial a} \frac{1}{x^{2}+a^{2}} d x=-\frac{1}{a} \tan ^{-1} \frac{x}{a} \times \frac{1}{a}+\frac{-x}{a\left(x^{2}+a^{2}\right)}$

$\int_{0}^{x} \frac{2 a^{2}}{x^{2}+a^{2}} d x=-\frac{1}{a} \tan ^{-1} \frac{x}{a} \times \frac{1}{a}+\frac{-x}{a\left(x^{2}+a^{2}\right)}$

$\int_{0}^{x} \frac{d x}{\left(x^{2}+a^{2}\right)^{2}} d x=\frac{1}{2 a^{3}} \tan ^{-1} \frac{x}{a}+\frac{x}{2 a^{2}\left(x^{2}+a^{2}\right)}$

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