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

290 Points
12 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,

290 Points
12 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,