Question

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

• A. $\log \left(3\right)$
• B. $\log \left(1\right)$
• C. $\log \left(4\right)$
• D. $\log \left(2\right)$

Answer 1

The correct option is C

$\log \left(4\right)$

Explanation for the correct option:

Evaluating the given integral:

Given integral is $\mathrm{I}={\int }_0^1\frac{1}{\mathrm{x}+\sqrt{\mathrm{x}}}\mathrm{dx}.....\left(1\right)$

Substituting $\sqrt{\mathrm{x}}=\mathrm{t}$

Differentiate with respect to $\mathrm{x}$ we get,
$$\begin{gathered} \frac{1}{2\sqrt{\mathrm{x}}}\mathrm{dx}=\mathrm{dt} \\ \Rightarrow \mathrm{dx}=2\sqrt{\mathrm{x}}\mathrm{dt} \\ \Rightarrow \mathrm{dx}=2\mathrm{tdt} \end{gathered}$$

Substituting this in equation $\left(1\right)$ we get
$\begin{array}{rcl}\mathrm{I}& =& {\int }_0^1\frac{2\mathrm{tdt}}{{\mathrm{t}}^2+\mathrm{t}}\left[\begin{array}{l}\because \sqrt{x}=t,\\ \Rightarrow x=t^2\end{array}\right]\\ & =& 2{\int }_0^1\frac{\mathrm{t}d\mathrm{t}}{\mathrm{t}(\mathrm{t}+1)}\end{array}$

$\begin{array}{rcl}& =& 2{\int }_0^1\frac{\mathrm{dt}}{1+\mathrm{t}}\\ & =& 2\left[\log \right(1+\mathrm{t}){]}_0^1\left[\because \int \frac{1}{x}\mathrm{dx}=\log \left|x\right|\right]\\ & =& 2\left[\log \right(1+1)-\log (1+0\left)\right]\end{array}$

$=2\left(\log \right(2)-\log (1\left)\right)$
$=2\log \left(2\right)$
$=\log {\left(2\right)}^2\left[\because \mathrm{nlog}\left(m\right)=\log {\left(m\right)}^n\right]$
$=\log \left(4\right)$

Therefore, the value of the given integral is equal to $\log \left(4\right)$.

Hence, option (C) is the correct option.