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