Question

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

The correct option is D $-{\left(\frac{x^4+1}{x^4}\right)}^{\frac{1}{4}}+c$

$I=\int \frac{dx}{x^2(x^4+1{)}^{\frac{3}{4}}}=\int \frac{dx}{x^5(1+\frac{1}{x^4}{)}^{\frac{3}{4}}}$
Let $t=1+\frac{1}{x^4}$
$\Rightarrow dt=-\frac{4}{x^5}dx$
$\therefore I=-\frac{1}{4}\int \frac{dt}{t^{3/4}}$
$I=-t^{1/4}+C$
$\Rightarrow I=-{\left(\frac{x^4+1}{x^4}\right)}^{\frac{1}{4}}+C$