Question

If $\int \frac{\sqrt{x^2+1}[\log (x^2+1)-2\log x]}{x^4}dx$ simplies to $\frac{\sqrt{(x^2+1{)}^3}[a\log (x^2+1)+b\log x+c]}{{px}^3}$, then find the value of $p-(a+b+c)$.

Answer 1

$\int \frac{\sqrt{x^2+1}(\log (x^2+1)-2\log x}{x^4}$

$=\int \frac{\frac{\sqrt{x^2+1}}{x}(\log \left(\frac{x^2+1}{x^2}\right)}{x^3}dx$

$=\sqrt{1+\frac{1}{x^2}}\log 1+\frac{1}{x^2}\times \frac{dx}{x^3}$

Let $1+\frac{1}{x^2}=t$

$\frac{-2}{x^3}dx=dt$

$\frac{dx}{x^3}=\frac{dt}{-2}$

$=\sqrt{t}\log t\times \frac{dt}{-2}$

$=\frac{-1}{2}\int \sqrt{t}\log tdt$

$=\frac{-1}{2}(\log t\times \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{t^{\frac{3}{2}}}{\frac{3}{2}}\times \frac{1}{t}dt)$

$=\frac{-1}{2}(\frac{2}{3}\log t\times t^{\frac{3}{2}}-\frac{2}{3}\times \frac{t^{\frac{3}{2}}}{\frac{3}{2}})$

$=\frac{-1}{3}{t}^{\frac{3}{2}}(\log t-\frac{2}{3})$

$=\frac{-1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}(\log 1+\frac{1}{x^2}-\frac{2}{3})$

$=\frac{-1}{3}\times \frac{{(x+1)}^{\frac{3}{2}}\times (3\log {(x}^2+1)-6\log x-2)}{3x^3}$

$=\frac{{(x+1)}^{\frac{3}{2}}\times (-3\log {(x}^2+1)+6\log x+2)}{9}$

$p-(a+b+c)=9-(-3+6+2)=4$