Question

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

• A. $\frac{4{(\sqrt{x}+1)}^{\frac{3}{2}}}{3}+c$
• B. $\frac{3{(\sqrt{x}+1)}^{\frac{4}{3}}}{4}+c$
• C. $\frac{{(\sqrt{x}+1)}^{\frac{4}{3}}}{4}+c$
• D. $\frac{{(\sqrt{x}+1)}^{\frac{2}{3}}}{3}+c$

Answer 1

Answer: (A)

$\frac{4{(\sqrt{x}+1)}^{\frac{3}{2}}}{3}+c$

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

$\sqrt{x}=t$

$d\sqrt{x}=dt$

$\frac{1}{2\sqrt{x}}dx=dt$

$\frac{dx}{\sqrt{x}}=2dt$

$\int \sqrt{1+t}.2dt$

$=2\int \sqrt{1+t}dt$

$=\frac{2(t+1{)}^{\frac{3}{2}}}{\frac{3}{2}}+c$

Put the value of 't'

$=\frac{4(\sqrt{x}+1{)}^{\frac{3}{2}}}{3}+c$

$\int \frac{\sqrt{1+\sqrt{x}}}{\sqrt{x}}dx=\frac{4(\sqrt{x}+1{)}^{\frac{3}{2}}}{3}+c$