Question

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

• A. $11/15$
• B. $14/15$
• C. $2/5$
• D. none of these

Answer 1

The correct option is D none of these

Let $I={\int }_0^3\frac{dx}{\sqrt{x+1}+\sqrt{5x+1}}={\int }_0^3\frac{\sqrt{x+1}-\sqrt{5x+1}}{-4x}dx=I_1+I_2$
Where $I_1=-\frac{1}{4}{\int }_0^3\frac{\sqrt{x+1}}{x}dx$
Put $x+1=t^2\Rightarrow dx=2tdt$
$I_1=-\frac{1}{4}{\int }_1^2\frac{2t^2}{t^2-1}dt=-\frac{1}{2}{\int }_1^2(-\frac{1}{2(t+1)}+\frac{1}{2(t-1)}+1)dt$
$=\frac{1}{4}{\left[\log (t+1)\right]}_1^2-\frac{1}{4}{\left[\log (t-1)\right]}_1^2+\frac{1}{2}{\left[t\right]}_1^2$
And $I_2=\frac{1}{4}{\int }_0^3\frac{\sqrt{5x+1}}{x}dx$
Put $5x+1=u^2$
$I_2=\frac{1}{10}{\int }_1^4\frac{u^2}{1+u^2}du=\frac{1}{2}\int (-\frac{1}{2(u+1)}+\frac{1}{2(u-1)}+1)du$
$=-\frac{1}{4}{\left[\log (u+1)\right]}_1^4+\frac{1}{4}{\left[\log (u-1)\right]}_1^4+\frac{1}{4}{\left[u\right]}_1^4$