Question

${\int }^{}\frac{{(x-x^5)}^{\frac{1}{5}}}{x^6}dx$ is equal to

• A. $\frac{5}{24}{\left(\frac{1}{x^4}-1\right)}^{\frac{6}{5}}+C$
• B. $\frac{5}{24}{\left(1-\frac{1}{x^4}\right)}^{\frac{6}{5}}+C$
• C. $-\frac{5}{24}{\left(\frac{1}{x^4}-1\right)}^{\frac{6}{5}}+C$
• D. none of these

Answer 1

Answer: (C)

$-\frac{5}{24}{\left(\frac{1}{x^4}-1\right)}^{\frac{6}{5}}+C$

$\int \frac{(x-x^5{)}^{1/5}}{x^6}dx$

$=\int \frac{x{(\frac{1}{x^4}-1)}^{1/5}}{x^6}dx$

$=\int \frac{{(\frac{1}{x^4}-1)}^{1/5}}{x^5}dx$

$\frac{1}{x^4}-1=t$

$=\frac{-4}{x^5}dx=dt$

$\frac{dx}{x^5}=-\frac{dt}{4}$

$=\int \frac{-t^{1/5}dt}{4}$

$=\frac{-5}{24}{t}^{6/5}+c$

$=\frac{-5}{24}{(\frac{1}{x^4}-1)}^{6/5}+c$

C is correct.