# Integrate the function$\int\frac{\sqrt{x^2+1}[log(x^2+1)-2logx]}{x^4}$

Toolbox:
• (i)If an integral function $f(x)=t,$ then $f'(x)dx=dt$ hence the integral function is $\int f(x)dx=\int t.dt$
• (ii)$log a-lod b=log |a/b|+c$
• (iii)$2 log x=logx^2$
• (iv) $\int udv=uv-\int vdu$
Given $I=\int\frac{\sqrt{x^2+1}[log(x^2+1)-2logx]}{x^4}dx$

Consider $\frac{\sqrt{x^2+1}[log(x^2+1)-2logx]}{x^4}$

We can write this as

$=\frac{\sqrt{x^2+1}}{x^4} \bigg[log (x^2+1)-log x^2 \bigg]\qquad(2 log x = log x^2)$

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

Hence $=\frac{\sqrt{x^2+1}}{x^4} \bigg[log \frac{x^2+1}{x^2} \bigg]$

$=\frac{\sqrt{x^2+1}}{x^4} \bigg[log (1+\frac{1}{x^2}) \bigg]$

$=\frac{1}{x^3}\sqrt{\frac{x^2+1}{x^2}} \bigg[log (1+\frac{1}{x^2}) \bigg]$

$=\frac{1}{x^3}\sqrt{1+\frac{1}{x^2}} \bigg[log (1+\frac{1}{x^2}) \bigg]$

$I=\int\frac{1}{x^3}\sqrt{1+\frac{1}{x^2}} \bigg[log (1+\frac{1}{x^2}) \bigg]dx$

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

On differentiating w.r.t. x we get

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

Now substituting t and dt,

$I=-\frac{1}{2}\int \sqrt t \;log t\; dt$

This is of the form $\int udv$

we know $\int udv=uv-\int vdu$

Let $u=log t,$differentiating w.r.t t

$du=\frac{1}{t} dt$

Let $dv=t^{1/2}dt.$ on integrating

$v=\frac{t^{3/2}}{3/2}=\frac{2}{3}t^{3/2}$

Substituting for u,v,du and dv we get,

Substituting for t we get,

$\frac{1}{3}(1+\frac{1}{x^2})^{3/2} log(1+\frac{1}{x^2})-\frac{2}{9}(1+\frac{1}{x^2})^{3/2}+c$

$=\frac{1}{3}\bigg[(1+\frac{1}{x^2})^{3/2} log(1+\frac{1}{x^2})-\frac{2}{3}(1+\frac{1}{x^2})^{3/2}\bigg]+c$

$I=\frac{1}{3}(1+\frac{1}{x^2})^{3/2} \bigg[ log(1+\frac{1}{x^2})-\frac{2}{3}\bigg]+c$