Question

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

• A. ${(1+\frac{1}{x^4})}^{1/4}+c$
• B. $-{(1+\frac{1}{x^4})}^{1/4}+c$
• C. ${(1+\frac{1}{x^4})}^{1/2}+c$
• D. $-{(1+\frac{1}{x^4})}^{3/4}+c$

Answer 1

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

$\int \frac{{(x^4+1)}^{\frac{3}{4}}}{x^2}dx$
$=\int \frac{1}{x}{(1+\frac{1}{x^4})}^{3/4}dx$
$=-\frac{1}{4}\int \frac{1}{t^{3/4}}dt=-\frac{1}{4}\frac{t^{3/4}}{1/4}+c=t^{1/4}+c$
where $t=1+\frac{1}{x^4}$
$=-{(1+\frac{1}{x^4})}^{1/4}+c$
Hence, option 'B' is correct.