Question

Solve: ${\int }_0^1\frac{1}{x+\sqrt{x}}dx$

• A. $\log 2$
• B. $2\log 4$
• C. $\log 4$
• D. None of these

Answer 1

The correct option is C $\log 4$

${\int }_0^1\frac{1}{x+\sqrt{x}}dx$

Let,

$u=\sqrt{x}$

$⟹u^2=x$

and, $du=\frac{1}{2\sqrt{x}}=\frac{1}{2u}du$

$2udu=dx$

$={\int }_0^1\frac{1}{u^2+u}2udu$

$=2{\int }_0^1\frac{u}{u(u+1)}du$

Let, $t=u+1$

$dt=du$

$=2{\int }_0^2\frac{1}{t}dt$

$=2[\ln t{]}_0^2$

$=\left[2\ln \right(1+u){]}_0^1$

$=2\ln 2-2\ln 1$

$=2\ln 2$