Question

$\int \frac{1}{x^2(x^4+1{)}^{\frac{3}{4}}}$ dx is equal to __________

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

Answer 1

The correct option is A

$\frac{-(1+x^4{)}^{\frac{1}{4}}}{x}+C$

$\frac{1}{x^2(x^4+1{)}^{\frac{3}{4}}}=\frac{1}{x^5(1+\frac{1}{x^4}{)}^{\frac{3}{4}}}Let1+\frac{1}{x^4}=t\Rightarrow \frac{1}{x^5}dx=-\frac{1}{4}dt\therefore \int \frac{1}{x^2(x^4+1{)}^{\frac{3}{4}}}dx=\frac{-1}{4}\int \frac{1}{t^{\frac{3}{4}}}dt=\frac{-(x^4+1{)}^{\frac{1}{4}}}{x}$