Question

$\int \frac{\sqrt{1+x^{2n}}\{\log (1+x^{2n})-2n\log x\}}{x^{3n+1}}dx=-\frac{1}{2n}\int \sqrt{t}.\log tdt=-\frac{1}{2n}[\frac{2}{3}t\sqrt{t}\log t-\frac{4}{9}t\sqrt{t}]$ where $t=1+\frac{1}{x^{2n}}$. If this is true enter 1, else enter 0.

Answer 1

Let $I=\int \frac{\sqrt{1+x^{2n}}\{\log (1+x^{2n})-2n\log x\}}{x^{3n+1}}dx$
$=\int \log \left\{\frac{1+x^{2n}}{x^{2n}}\right\}\cdot \frac{\sqrt{1+x^{2n}}}{x^n.x^{2n+1}}dx$
Put $\frac{1+x^{2n}}{x^{2n}}=1+\frac{1}{x^{2n}}=t\Rightarrow -\frac{2n}{x^{2n+1}}dx=dt$
Therefore
$I=-\frac{1}{2n}\int \sqrt{t}.\log tdt=-\frac{1}{2n}[\frac{2}{3}t\sqrt{t}\log t-\frac{4}{9}t\sqrt{t}]$