Question

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

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

Answer 1

The correct option is D $-{(1+\frac{1}{x^4})}^{1/4}+C$

$\int \frac{dx}{x^2(x^4+1{)}^{3/4}}=\int \frac{dx}{x^2\cdot x^3{\left(\frac{x^4+1}{x^4}\right)}^{3/4}}=\int \frac{dx}{x^5{(1+\frac{1}{x^4})}^{3/4}}$
Put $1+\frac{1}{x^4}=t\Rightarrow -\frac{4}{x^5}dx=dt$
So, integral is
$I=-\frac{1}{4}\int \frac{dt}{t^{3/4}}=-\frac{1}{4}\frac{(t{)}^{\frac{1}{4}}}{\frac{1}{4}}+c=-t^{\frac{1}{4}}+c=-{(1+\frac{1}{x^4})}^{1/4}+c$

Hence, option D is correct.