Question

$\int \frac{x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}dx$

Answer 1

Consider the given integral.

$I=\int \frac{x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}dx$

$I=\int \frac{x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}\times \frac{\sqrt{x^2+a^2}-\sqrt{x^2-a^2}}{\sqrt{x^2+a^2}-\sqrt{x^2-a^2}}dx$

$I=\int \frac{x(\sqrt{x^2+a^2}-\sqrt{x^2-a^2})}{2a^2}dx$

$I=\int \frac{x\sqrt{x^2+a^2}}{2a^2}dx-\int \frac{\sqrt{x^2-a^2}}{2a^2}dx$

$I=I_1+I_2$ ......(1)

Therefore,

$I_1=\int \frac{x\sqrt{x^2+a^2}}{2a^2}dx$

Put

$t=x^2+a^2$

$dt=2xdx$

Therefore,

$I_1=\frac{1}{4a^2}\int \sqrt{t}dx$

$I_1=\frac{1}{4a^2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$

$I_1=\frac{1}{6a^2}{(x^2+a^2)}^{\frac{3}{2}}+C$

Similarly,

$I_2=\int \frac{x\sqrt{x^2-a^2}}{2a^2}dx$

Put

$t=x^2-a^2$

$dt=2xdx$

Therefore,

$I_2=\frac{1}{4a^2}\int \sqrt{t}dx$

$I_2=\frac{1}{4a^2}\left[\frac{t^{\left(\frac{3}{2}\right)}}{\frac{3}{2}}\right]+C$

$I_2=\frac{1}{6a^2}{(x^2-a^2)}^{\frac{3}{2}}+C$

Therefore, substituting values if $I_1$ and $I_2$ in equation (1),

$I=\frac{1}{6a^2}{(x^2+a^2)}^{\frac{3}{2}}-\frac{1}{6a^2}{(x^2-a^2)}^{\frac{3}{2}}+C$

$I=\frac{1}{6a^2}[{(x^2+a^2)}^{\frac{3}{2}}-{(x^2-a^2)}^{\frac{3}{2}}]+C$

Hence, this is the value of above integral.