Question

$\int \frac{(x+x^3{)}^{1/3}}{x^4}dx$ is equal to

• A. $\frac{3}{8}{(\frac{1}{x^2}-1)}^{4/3}+C$
• B. $-\frac{3}{8}\left[{(1+\frac{1}{x^2})}^{4/3}\right]+C$
• C. $-\frac{1}{8}{(1+\frac{1}{x^2})}^{4/3}+C$
• D. $\frac{1}{8}{(\frac{1}{x^2}-1)}^{4/3}+C$

Answer 1

Answer: (B)

$-\frac{3}{8}\left[{(1+\frac{1}{x^2})}^{4/3}\right]+C$

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

$=\int \frac{{\left[x^3(\frac{1}{x^2}+1)\right]}^{\frac{1}{3}}}{x^4}dx$

$=\int \frac{x{\left[(\frac{1}{x^2}+1)\right]}^{\frac{1}{3}}}{x^4}dx$

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

Let $\frac{1}{x^2}+1=t$

then,

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

$I=\int \frac{-1}{2}(t{)}^{\frac{1}{3}}dt$

$=\frac{-1}{2}\times \frac{3}{4}(t{)}^{\frac{4}{3}}+c$

$=\frac{-3}{8}{(\frac{1}{x^2}+1)}^{\frac{4}{3}}+c$