\( \int\displaystyle\frac{1+x+\sqrt{x+x^2}}{\sqrt{1+x}+\sqrt{x}}\,dx \) is equal to (Where \( c \) is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{2}{3}(1+x)^{\displaystyle\frac{3}{2}}+c \)
Since \( \sqrt{x+x^2}=\sqrt{x}\sqrt{1+x} \), the numerator factorises:
\( \displaystyle 1+x+\sqrt{x}\sqrt{1+x}=\sqrt{1+x}\left(\sqrt{1+x}+\sqrt{x}\right) \)
The bracket cancels with the denominator:
\( \displaystyle \int\frac{1+x+\sqrt{x+x^2}}{\sqrt{1+x}+\sqrt{x}}\,dx=\int\sqrt{1+x}\,dx \)
\( \displaystyle =\frac{2}{3}(1+x)^{3/2}+c \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{3}{2}(1+x)^{\displaystyle\frac{3}{2}}+c \) — Incorrect. The coefficient is inverted: integrating \( u^{1/2} \) gives \( \frac{2}{3}u^{3/2} \).
Option (3) \( \displaystyle\frac{2}{3}(1+x)^{\displaystyle\frac{3}{2}}+\displaystyle\frac{2}{3}x^{\displaystyle\frac{2}{3}}+c \) — Incorrect. The integrand reduces to \( \sqrt{1+x} \) alone, so there is no extra term in x.
Option (4) \( \displaystyle\frac{2}{3}\displaystyle\frac{1}{(1+x)^{\displaystyle\frac{3}{2}}}+c \) — Incorrect. The power of \( (1+x) \) is \( +\frac{3}{2} \), not \( -\frac{3}{2} \).