Question

The value of $\int \frac{dx}{x^2(x^4+1{)}^{3/4}}$ is

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

Answer 1

Answer: (D)

$\int \frac{dx}{x^2(x^4+1{)}^{3/4}}=\int \frac{dx}{x^5(1+\frac{1}{x^4}{)}^{3/4}}$
put $1+\frac{1}{x^4}=t^4\Rightarrow \frac{-4}{x^5}dx=4t^{3}dt\Rightarrow \frac{dx}{x^5}=-t^{3}dt$
Hence, the integral becomes
$\int \frac{-t^{3}dt}{t^3}=-\int dt=-t+c=-(1+\frac{1}{x^4}{)}^{1/4}+c$