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integrate {[(x 2 +1) 1/2 [ln(x 2 +1) - 2lnx]]/x 4 } w.r.t x

integrate  {[(x2+1)1/2[ln(x2+1) - 2lnx]]/x4} w.r.t x

Grade:12th Pass

2 Answers

Ashwin Muralidharan IIT Madras
290 Points
10 years ago

Hi Suchita,

 

Take x2 from the root, and combine the two log, you will have {(1+1/x2)1/2*ln(1+1/x2)}/x3 dx.

Now make the substitution 1+1/x^2 = t, and you have dx/x^3 = -3dt.

And the integral will be ∫(-3)*t*lnt dt, which is directly integrated by parts

= -3/2 { t*2 lnt - ∫tdt }. And that solves the problem, substitute back for t = 1 + 1/x^2.

 

Hope that helps,

 

All the best,

Regards,

Ashwin (IIT Madras).

Ashwin Muralidharan IIT Madras
290 Points
10 years ago

Hi Suchita,

 

Please note, here again the method of integration is the same as mentioned above.

 

Where 2lnx = lnx2

And ln(1+x2) - lnx2 = ln[(1+x2)/x2] = ln(1+1/x2)

 

And when x^2, is taken from the square root term, it cancels one x in the Dr, leaving x3.

 

So sub 1+1/x2 = t. Which will give dx/x3 = -dt/2

 

Please be careful with such calculations (previously there was a mistake while doing it very fast, and hence a mistake in the constant term. Please do take time while practicing questions to not make such very silly mistake).

 

Hope that helps.

 

All the best.

Regards,

Ashwin (IIT Madras).

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