AP EAMCET · Maths · Indefinite Integration
\(\int \sqrt{x+\sqrt{x^2+2}} d x=\)
- A \(\frac{3}{2}(x+\sqrt{x+2})^{\frac{3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{1}{4}}+C\)
- B \(\frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{1}{4}}+C\)
- C \(\left(x+\sqrt{x^2+2}\right)^{\frac{-3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{-1}{2}}+C\)
- D \(\frac{\left(x+\sqrt{x^2+2}\right)^2-6}{3 \sqrt{x+\sqrt{x^2+2}}}+C\)
Answer & Solution
Correct Answer
(D) \(\frac{\left(x+\sqrt{x^2+2}\right)^2-6}{3 \sqrt{x+\sqrt{x^2+2}}}+C\)
Step-by-step Solution
Detailed explanation
\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}…
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