∫(4x5-7x4+8x3-2x2+4x-7)/x2(x2+1)2 dx

Hi Suchita,

Let's see how to reduce this to integrable form, step by step

First, note that x=1 is a root for the Nr.

So Nr can be factorised as (x-1)(4x⁴-3x³+5x²+3x+7).

Now note that the terms in the second bracket can be written as [4(x⁴+2x²+1)-3x³-3x²+3x+3] = [4(x²+1)²-3(x³+x²-x-1)].

So now split into two integrals.

I1 = 4(x-1)/x²------- (directly integrated by using x^n formula, as it is 1/x - 1/x² form)

I2 = -3(x-1)[x³+x²-x-1]/x²(x²+1)².

for I2, again note that x=1 is a root for the cubic exp, hence it could be factorised as (x-1)(x²+2x+1) = (x-1)(x+1)².

So the integral is of the form (ignoring the constants) ------- [(x-1)²(x+1)²]/[x²(x²+1)²] = (x²-1)²/x²(x²+1)².

For this make the sub 1/x = t, so that dx/x² = -dt (this is done to eliminate the x² in the Dr.)

So you have an integral of the form (1-t²)²/(1+t²)² = [(1+t²)² - 4t²]/(1+t²)².

Now split this into two integrals.

First one is ∫1.dt

Second one is of the form ∫t²/(t⁴+2t²+1) --------- (standard integral, integrated by dividing Nr and Dr by t²)

which reduces to 1/(t²+1/t²+2) = (1/2)* [(1+1/t²)+(1-1/t²)]/(t²+1/t²+2).... for 1st integral t+1/t=y and for 2nd t-1/t=y.

And hence that standard integral is integrated.

And that solves the problem.

Also please note, for another integration question you'd posted on the forum, a small change in constant is made and solution is re-posted. Have a re-look at that one too.

Hope that helps.

All the best,

Regards,

Ashwin (IIT Madras).

Approved