Question

Evaluate \[ \int \frac{1+x+\sqrt{x+x^{2}}}{\sqrt{1+x}+\sqrt{x}}dx \]

• A. $\frac{3}{2}(1+x)^{\frac{3}{2}}+c$
• B. $\frac{2}{3}(1+x)^{\frac{3}{2}}+c$
• C. $\frac{2}{3}(1+x)^{\frac{3}{2}}+\frac{2}{3}x^{\frac{2}{3}}+c$
• D. $\frac{2}{3}\frac{1}{(1+x)^{\frac{3}{2}}}+c$

Hint:

Whenever $\sqrt{x(1+x)}$ appears, try factoring into $\sqrt{x}$ and $\sqrt{1+x}$ before substitution.

Answer 1

The Correct Option is B

Concept: We simplify radicals using: \[ x+x^2=x(1+x), \quad \sqrt{x+x^2}=\sqrt{x}\sqrt{1+x} \]

Step 1: {Rewrite numerator fully.}
\[ 1+x+\sqrt{x}\sqrt{1+x} \]

Step 2: {Rewrite denominator.}
\[ \sqrt{1+x}+\sqrt{x} \]

Step 3: {Let substitution be $t=\sqrt{1+x}-\sqrt{x}$.}
Then: \[ t(\sqrt{1+x}+\sqrt{x})=1 \]

Step 4: {Simplify integrand using rationalization identity.}
The expression reduces to: \[ \frac{d}{dx}\left(\frac{2}{3}(1+x)^{3/2}\right) \]

Step 5: {Integrate directly.}
\[ \int = \frac{2}{3}(1+x)^{3/2}+c \]