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TS EAMCET · Maths · Indefinite Integration

\(\int \frac{x^3}{\sqrt{1+x^2}} d x\) is equal to

  1. A \(\sqrt{1+x^2}-\frac{x}{3}\left(1+x^2\right)^{3 / 2}+C\)
  2. B \(x \sqrt{1+x^2}+\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C\)
  3. C \(x^2 \sqrt{1+x^2}-\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C\)
  4. D \(x^2 \sqrt{1+x^2}-\frac{1}{3}\left(1+x^2\right)^{1 / 2}+C\)
Verified Solution

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…