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

1 Answer

  • (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
Substituting for u,v,du and dv we get,
Substituting for t we get,
$\frac{1}{3}(1+\frac{1}{x^2})^{3/2} log(1+\frac{1}{x^2})-\frac{2}{9}(1+\frac{1}{x^2})^{3/2}+c$
$=\frac{1}{3}\bigg[(1+\frac{1}{x^2})^{3/2} log(1+\frac{1}{x^2})-\frac{2}{3}(1+\frac{1}{x^2})^{3/2}\bigg]+c$
$I=\frac{1}{3}(1+\frac{1}{x^2})^{3/2} \bigg[ log(1+\frac{1}{x^2})-\frac{2}{3}\bigg]+c$


answered Feb 18, 2013 by meena.p