Question

$\int \sqrt{x^6+1}.\frac{\log (x^6+1)-6\log x}{x^{10}}dx=\frac{1}{6}[\frac{2}{3}{t}^{3/2}\log t-\frac{2}{3}\int t^{3/2}\frac{1}{t}]$ where $t=1+\frac{1}{x^6}$. If this is true enter 1, else enter 0.

Answer 1

Let $I=\int \sqrt{x^6+1}.\frac{\log (x^6+1)-6\log x}{x^{10}}dx$

Write $\frac{1}{x^{10}}=\frac{1}{x^7.x^3}=\frac{1}{x^7.\sqrt{x^6}}$

We get
$I=\int \sqrt{\frac{x^6+1}{x^6}}\log \frac{x^6+1}{x^6}\cdot \frac{1}{x^7}dx$

Put $\frac{x^6+1}{x^6}=1+\frac{1}{x^6}=t\Rightarrow \frac{-6}{x^7}dx=dt$

Therefore
$I=\int -\frac{1}{6}\sqrt{t}\log tdt=\frac{-1}{6}[\frac{2}{3}{t}^{3/2}\log t-\frac{2}{3}\int t^{3/2}\frac{1}{t}]dt$
$=-\frac{1}{6}[\frac{2}{3}{t}^{3/2}\log t-\frac{2}{3}\int t^{3/2}\frac{1}{t}]$