Question

The value of the integral ${\int }_0^3\frac{dx}{\sqrt{x+1}+\sqrt{5x+1}}dxis$

• A. $\frac{11}{15}$
• B. $\frac{14}{15}$
• C. $\frac{2}{5}$
• D. $1+\frac{1}{2}log\left(\frac{3}{5}\right)$

Answer 1

The correct option is D $1+\frac{1}{2}log\left(\frac{3}{5}\right)$

$Put5x+1=y^2,x+1=t^2.Thendx=\frac{1}{5}2ydy,dx=2tdt$
${\int }_0^3\frac{1}{\sqrt{x+1}+\sqrt{5x+1}}dx={\int }_0^3\frac{\sqrt{5x+1}-\sqrt{x+1}}{4x}dx=\frac{1}{4}\left[{\int }_0^3\frac{\sqrt{5x+1}}{x}dx-{\int }_0^3\frac{\sqrt{x+1}}{x}dx\right]$
$=\frac{1}{4}\left[{\int }_1^4\frac{\frac{y}{y^2-1}}{5}.\frac{2y}{5}dy-{\int }_1^2\frac{t}{t^2-1}.2tdt\right]=\frac{1}{2}\left[{\int }_1^4\frac{y^2}{y^2-1}dy-{\int }_1^2\frac{t^2}{t^2-1}dt\right]$
$=\frac{1}{2}\left[{\int }_1^4(1+\frac{1}{y^2-1})dy-{\int }_1^2(1+\frac{1}{t^2-1})dt\right]$
$=\frac{1}{2}\{{[y+\frac{1}{2}log\left|\frac{y-1}{y+1}\right|]}_1^4-{[t+\frac{1}{2}log\left|\frac{t-1}{t+1}\right|]}_1^4\}$
$=\frac{1}{2}[4+\frac{1}{2}log\left(\frac{3}{5}\right)-1-2-\frac{1}{2}-log\left(\frac{1}{3}\right)+1]=\frac{1}{2}[2+\frac{1}{2}log\frac{9}{5}]$
$=1+\frac{1}{2}log\left(\frac{3}{5}\right)$