ExamBro
ExamBro
TS EAMCET · Maths · Indefinite Integration

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

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

Answer & Solution

Correct Answer

(B) \(\frac{2}{3}(1+x)^{3 / 2}+C\)

Step-by-step Solution

Detailed explanation

\(\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x\) \(=\int \frac{(1+x)+\sqrt{x(1+x)}}{\sqrt{x}+\sqrt{1+x}} d x\) \(=\int \frac{\sqrt{1+x}(\sqrt{1+x}+\sqrt{x})}{\sqrt{x}+\sqrt{1+x}} d x\) \(=\int \sqrt{1+x} d x=\frac{(1+x)^{3 / 2}}{3 / 2}+C\) \(=\frac{2}{3}(1+x)^{3 / 2}+C\)