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∫(4x 5 -7x 4 +8x 3 -2x 2 +4x-7)/x 2 (x 2 +1) 2 dx

∫(4x5-7x4+8x3-2x2+4x-7)/x2(x2+1)2 dx

Grade:12th Pass

1 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

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,


Ashwin (IIT Madras).


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