Question

$\int \frac{dx}{x+\sqrt{x}}$ is equal to

• A. $\frac{1}{2}\log (1+\sqrt{x})+c$
• B. $2\log (1+\sqrt{x})+c$
• C. $\frac{1}{4}\log (1+\sqrt{x})+c$
• D. $3\log (1+\sqrt{x})+c$

Answer 1

The correct option is B $2\log (1+\sqrt{x})+c$

Let $I=\int \frac{dx}{x+\sqrt{x}}=\int \frac{1}{\sqrt{x}(1+\sqrt{x})}dx$

Put $1+\sqrt{x}=t$

$\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$

$\therefore I=2\int \frac{1}{t}dt=2\log t+c$

$I=2\log (1+\sqrt{x})+c$