Question

$\int x^{-\frac{2}{3}}{(1+x^{\frac{1}{2}})}^{-\frac{5}{3}}dx$ is equal to

• A. $3{(1+x^{-\frac{1}{2}})}^{-\frac{1}{3}}+c$
• B. $3{(1+x^{-\frac{1}{2}})}^{-\frac{2}{3}}+c$
• C. $3{(1+x^{\frac{1}{2}})}^{-\frac{2}{3}}+c$
• D. none of these

Answer 1

The correct option is A $3{(1+x^{-\frac{1}{2}})}^{-\frac{1}{3}}+c$

$\int x^{-\frac{2}{3}}(1+x^{\frac{1}{2}}{)}^{-\frac{5}{3}}dx$

$=\int x^{\frac{-2}{3}}\left(x^{\frac{1}{2}}\right(1+x^{-\frac{1}{2}}){)}^{\frac{-5}{3}}dx$

$=\int x^{-\frac{2}{3}}.x^{\frac{-5}{6}}(1+x^{\frac{-1}{2}}{)}^{\frac{-5}{3}}dx$

$=\int x^{\frac{-3}{2}}(1+x^{\frac{-1}{2}}{)}^{\frac{-5}{3}}dx$

put it $x^{\frac{-1}{2}}=t$

$x^{\frac{-1}{2}}{x}^{\frac{-3}{2}}dx=dt$

$=\int -2t^{\frac{-5}{3}}dt$

$=-2\times \frac{t^{\frac{-5}{3}}+1}{\frac{-5}{3}+1}+1$

$=3(1+x^{\frac{-1}{2}}{)}^{-\frac{2}{3}}+C$