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COMEDK · Maths · 28. Indefinite Integration

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

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Let \(I = \int \dfrac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} dx\). Observe that the numerator can be written as \(1+x+\sqrt{x}\sqrt{1+x} = \sqrt{1+x}(\sqrt{1+x}+\sqrt{x})\). Substituting this into the integral, we get…