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

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

Now $-\int \frac{1}{4t^{\frac{3}{4}}}dt=t^{\frac{1}{4}}+C=-\frac{(1+x^4{)}^{\frac{1}{4}}}{x}+C$