Question

The value of the integral $\underset{1}{\overset{16}{\int }}\frac{dx}{\sqrt{x}+\sqrt[4]{x}}$ is

• A. $4\ln \frac{3}{2}$
• B. $4\ln \frac{3}{2}+1$
• C. $4\ln \frac{3}{2}+2$
• D. none

Answer 1

Answer: (C)

Put $x=t^4\Rightarrow dx=4t^{3}dt$
$I=\underset{1}{\overset{2}{\int }}\frac{4t^3}{t^2+tt}=\underset{1}{\overset{2}{\int }}\frac{4t^2}{1+t}dt=4\left[\underset{1}{\overset{2}{\int }}\frac{\left(t^2-1\right)+1}{1+t}dt\right]=4\left[{\int }_1^2(t-1)dt+\underset{1}{\overset{2}{\int }}\frac{dt}{1+t}\right]$
$=4\left[{\left(\frac{t^2}{2}-t\right)}_1^2+{\ln (1+t)|}_1^2\right]=4\left[(0)-\left(-\frac{1}{2}\right)+\ln 3-\ln 2\right]=\left[\frac{1}{2}+\ln \frac{3}{2}\right]=4\ln \frac{3}{2}+2$