Question

The integral $\int \frac{dx}{x^2(x^4+1{)}^{3/4}}$ equals

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

Answer 1

Answer: (D)

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

$I=\int \frac{1}{x^2{(x^4+1)}^{\frac{3}{4}}}dx$

$=\int \frac{1}{x^2\times x^3{(1+\frac{1}{x^4})}^{\frac{3}{4}}}dx$

$=\int \frac{1}{x^5{(1+\frac{1}{x^4})}^{\frac{3}{4}}}dx$

Let $1+\frac{1}{x^4}=t$

then

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

$\frac{1}{x^5}dx=\frac{-1}{4}dt$

$I=\int \frac{-1}{4{\left(t\right)}^{\frac{3}{4}}}dt$

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

$=-t^{\frac{1}{4}}+C$

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

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