Question

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

Answer 1

$$\begin{gathered} \int \frac{xdx}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}} \\ \mathrm{Let}{x}^2=t \\ \Rightarrow 2x=\frac{dt}{dx} \\ \Rightarrow xdx=\frac{dt}{2} \\ \mathrm{Now},\int \frac{xdx}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}} \\ =\frac{1}{2}\int \frac{dt}{\sqrt{t+a^2}+\sqrt{t-a^2}} \\ =\frac{1}{2}\int \frac{dt}{\left(\sqrt{t+a^2}+\sqrt{t-a^2}\right)}\times \frac{\left(\sqrt{t+a^2}-\sqrt{t-a^2}\right)}{\left(\sqrt{t+a^2}-\sqrt{t-a^2}\right)} \\ =\frac{1}{2}\int \frac{\left(\sqrt{t+a^2}-\sqrt{t-a^2}\right)}{\left(t+a^2\right)-\left(t-a^2\right)}dt \\ =\frac{1}{4a^2}\int {\left(t+a^2\right)}^{\frac{1}{2}}dt-\frac{1}{4a^2}\int {\left(t-a^2\right)}^{\frac{1}{2}}dt \\ =\frac{1}{4a^2}\left[\frac{{\left(t+a^2\right)}^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]-\frac{1}{4a^2}\left[\frac{{\left(t-a^2\right)}^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]+\mathrm{C} \\ =\frac{1}{6a^2}\left[{\left(t+a^2\right)}^{\frac{3}{2}}-{\left(t-a^2\right)}^{\frac{3}{2}}\right]+\mathrm{C} \\ =\frac{1}{6a^2}\left[{\left(x^2+a^2\right)}^{\frac{3}{2}}-{\left(x^2-a^2\right)}^{\frac{3}{2}}\right]+\mathrm{C} \end{gathered}$$