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

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).

Approved

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).