Question

Solve $\underset{b}{\overset{a}{\int }}\frac{x}{\sqrt{a^2+x^2}}dx$

Answer 1

Consider the given integral.

$I={\int }_b^a\frac{x}{\sqrt{a^2+x^2}}dx$ …….. (1)

Let $t=a^2+x^2$

$\frac{dt}{dx}=0+2x$

$\frac{dt}{2}=xdx$

Therefore,

$I=\frac{1}{2}{\int }_{a^2+b^2}^{2a^2}\frac{1}{\sqrt{t}}dt$

$I=\frac{1}{2}{\left[2\sqrt{t}\right]}_{a^2+b^2}^{2a^2}$

$I=\sqrt{2a^2}-\sqrt{a^2+b^2}$

Hence, this is the answer.