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TS EAMCET · Maths · Indefinite Integration

For \(x \geq 0, \int \sqrt{x^2+2 x d x}\) is equal to

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

Answer & Solution

Correct Answer

(D) \(\frac{x+1}{2} \sqrt{x^2+2 x}-\frac{1}{2} \cosh ^{-1}(x+1)+C\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text { Let } I=\int \sqrt{x^2+2 x} d x=\int \sqrt{(x+1)^2-1} d x \\ & =\frac{(x+1)}{2} \sqrt{(x+1)^2-1}-\frac{(1)^2}{2} \\ & \log \left|(x+1)+\sqrt{(x+1)^2}-1+C\right| \\ & =\frac{(x+1)}{2} \sqrt{x^2+2 x}-\frac{1}{2} \cosh ^{-1}(x+1)+C\end{aligned}\)