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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\int\frac{\sqrt{x^2+1}[log(x^2+1)-2logx]}{x^4}\]

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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,
 
$\frac{1}{3}(1+\frac{1}{x^2})^{3/2} log(1+\frac{1}{x^2})-\frac{2}{9}(1+\frac{1}{x^2})^{3/2}+c$
 
$=\frac{1}{3}\bigg[(1+\frac{1}{x^2})^{3/2} log(1+\frac{1}{x^2})-\frac{2}{3}(1+\frac{1}{x^2})^{3/2}\bigg]+c$
 
$I=\frac{1}{3}(1+\frac{1}{x^2})^{3/2} \bigg[ log(1+\frac{1}{x^2})-\frac{2}{3}\bigg]+c$

 

answered Feb 18, 2013 by meena.p
 
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