Question

If $I=\int \frac{dx}{(1+x^4{)}^{1/4}}$, then I equals

• A. $\frac{1}{2}{\tan }^{-1}\left(\frac{x}{(1+x^4{)}^{1/4}}\right)+C$
• B. $\frac{1}{4}\log \left|\frac{1-(1+x^4{)}^{1/4}}{1+(1+x^4{)}^{1/4}}\right|+C$
• C. $\frac{1}{2}{\tan }^{-1}\frac{(1+x^4{)}^{1/4}}{x}-\frac{1}{4}\log \left|\frac{1-(1+x^4{)}^{1/4}}{1+(1+x^4{)}^{1/4}}\right|+C$
• D. $\frac{1}{2}{\tan }^{-1}\frac{(1+x^4{)}^{1/4}}{x}-\frac{1}{4}\log \left|\frac{x-(1+x^4{)}^{1/4}}{x+(1+x^4{)}^{1/4}}\right|+C$

Answer 1

The correct option is D $\frac{1}{2}{\tan }^{-1}\frac{(1+x^4{)}^{1/4}}{x}-\frac{1}{4}\log \left|\frac{x-(1+x^4{)}^{1/4}}{x+(1+x^4{)}^{1/4}}\right|+C$

$I=\int -\frac{dx}{x{(x^{-4}+1)}^{1/4}}=\int \frac{x^{-5}dx}{x^{-4}{(x^{-4}+1)}^{1/4}}$
Put $x^{-4}+1=t^4$
$I=-\int \frac{t^3}{(t^4+1)t}dt=\frac{1}{2}\int (\frac{1}{t^2+1}-\frac{1}{1-t^2})dt=\frac{1}{2}{\tan }^{-1}t-\frac{1}{4}\log \left|\frac{1-t}{1+t}\right|+c$