TS EAMCET · Maths · Indefinite Integration
\(\int \frac{x^3}{\sqrt{1+x^2}} d x\) is equal to
- A \(\sqrt{1+x^2}-\frac{x}{3}\left(1+x^2\right)^{3 / 2}+C\)
- B \(x \sqrt{1+x^2}+\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C\)
- C \(x^2 \sqrt{1+x^2}-\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C\)
- D \(x^2 \sqrt{1+x^2}-\frac{1}{3}\left(1+x^2\right)^{1 / 2}+C\)
Answer & Solution
Correct Answer
(C) \(x^2 \sqrt{1+x^2}-\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { Let } I=\int \frac{x^3}{\sqrt{1+x^2}} d x \\ & =\int x^2 \cdot \frac{x}{\sqrt{1+x^2}} d x \\ & \end{aligned} \] By using integration by parts…
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