$\int \sqrt{x+\sqrt{x^2+2}} d x=$

$\int \sqrt{x+\sqrt{x^2+2}} d x=$
  1. $\frac{3}{2}(x+\sqrt{x+2})^{\frac{3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{1}{4}}+C$
  2. $\frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{1}{4}}+C$
  3. $\left(x+\sqrt{x^2+2}\right)^{\frac{-3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{-1}{2}}+C$
  4. $\frac{\left(x+\sqrt{x^2+2}\right)^2-6}{3 \sqrt{x+\sqrt{x^2+2}}}+C$

Solution

$ \begin{aligned} & \quad\left(\sqrt{x^2+2}+x\right)\left(\sqrt{x^2+2}-x\right)=2 \\ & I=\int \sqrt{x+\sqrt{x^2+2}} d x \\ & \text { On putting } \sqrt{x^2+2}+x=t \\ & \Rightarrow \quad \sqrt{x^2+2}-x=\frac{2}{t} \\ & \therefore \quad \sqrt{x^2+2}=\frac{t+\frac{2}{t}}{2} \end{aligned} $ On putting $\sqrt{x^2+2}+x=t$ $ \Rightarrow \quad \sqrt{x^2+2}-x=\frac{2}{t} $ $ \sqrt{x^2+2}=\frac{t+\frac{2}{t}}{2} $ $ \begin{aligned} & \text { Also }\left(\frac{x}{\sqrt{x^2+2}}+1\right) d x=d t . \\ & \Rightarrow \quad d x=\frac{t+\frac{2}{t}}{2 t} d t \\ & \therefore \quad I=\int \sqrt{t} \cdot\left(\frac{t+\frac{2}{t}}{2 t}\right) d t=\int\left(\frac{\sqrt{t}}{2}+t^{-3 / 2}\right) d t \\ & \quad=\frac{t^{3 / 2}}{3}-\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \end{aligned} $ $\begin{aligned} & =\frac{t^2-6}{\frac{1}{2}}+C \\ \Rightarrow & \frac{\left(\sqrt{x^2+2}+x\right)^2-6}{3 \sqrt{\sqrt{x^2+2}+x}}+C\end{aligned}$

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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