Question

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

Answer 1

`∫   {x   dx}/{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2 }`
\[\text{Let x}^2 = t\]
\[ \Rightarrow 2x = \frac{dt}{dx}\]
\[ \Rightarrow \text{x dx }= \frac{dt}{2}\]
Now, `∫   {x   dx}/{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2 }`
\[ = \frac{1}{2}\int\frac{dt}{\sqrt{t + a^2} + \sqrt{t - a^2}}\]
\[ = \frac{1}{2}\int\frac{dt}{\left( \sqrt{t + a^2} + \sqrt{t - a^2} \right)} \times \frac{\left( \sqrt{t + a^2} - \sqrt{t - a^2} \right)}{\left( \sqrt{t + a^2} - \sqrt{t - a^2} \right)}\]
\[ = \frac{1}{2}\int\frac{\left( \sqrt{t + a^2} - \sqrt{t - a^2} \right)}{\left( t + a^2 \right) - \left( t - a^2 \right)}dt\]
\[ = \frac{1}{4 a^2}\int \left( t + a^2 \right)^\frac{1}{2} dt - \frac{1}{4 a^2}\int \left( t - a^2 \right)^\frac{1}{2} dt\]
\[ = \frac{1}{4 a^2}\left[ \frac{\left( t + a^2 \right)^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] - \frac{1}{4 a^2}\left[ \frac{\left( t - a^2 \right)^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = \frac{1}{6 a^2}\left[ \left( t + a^2 \right)^\frac{3}{2} - \left( t - a^2 \right)^\frac{3}{2} \right] + C\]
\[ = \frac{1}{6 a^2}\left[ \left( x^2 + a^2 \right)^\frac{3}{2} - \left( x^2 - a^2 \right)^\frac{3}{2} \right] + C\]