Expert: Scotto - 2/26/2011

Question
Determine the value of ∫ from 1 to 0 du/[(1 + u^2)(1+ u). I have no idea how to do partial fractions of this type and am really struggling. Your help will be most appreciated

Answer
I believe by that you mean ∫(1 downto 0)1/[(1 + u²)(1+u)] du.
Most of the time what is to the right of the du is not integrated, but I assume it is here.

To do this by partial fractions would be to let it be ∫(Au+B)/(1+u²) du + ∫C/(1+u) du.
When they are combined, the result is [(Au+B)(1+u) + C(1+u²)]/[(1+u²)(1+u)].
The result is [(A+C)u² + (A+B)u + (B+C)]/[(1+u²)(1+u)].

Looking back at the original problem, this says that (A+C)u² + (A+B)u + (B+C) = 1.
This means that A+C=0, A+B=0, and B+C=1.

Since A+C=0 says that A=-C, putting this in A+B=0 give -C+B=0.
Adding that to B+C=1 gives 2B=1, s0 B=1/2.
Since A+B=0, A=-1/2.  Since A+C=0, C=1/2.

This changes the problem to (1/2)∫(-u+1)/(u²+1) + 1/(u+1) du.

Now the last term, ∫(1/2)∫1/(u+1) du, integrates to (1/2)ln(u+1) = ln(u+1)/2.
This leaves (1/2)∫(-u+1)/(u²+1) du.
Multiplying by (-2)/(-2) gives [-1/4][∫2u/(u²+1) du + ∫2/(u²+1)du].

For the first integral, let v=u², so dv = 2u du, and that become ∫1/(v+1) dv,
and that is ln(v+1).  Looking back and inclucing the -1/4 out in front,
that gives (-1/4)ln(u²+1).

That leaves us with (-1/4)∫2/(u²+1) du, which is (-1/2)∫1/(u²+1) du.
Let the far side of a triangle be u, the near side be 1,  and so the hypotenuse is √(1+u²),
with an angle v.  This means that tan v = u, so sec²v dv = du.

Since sec²v = 1+u², sec²v dv = du can be divided by this.  That gives dv = [1/(u²+1)] du.
This reduces the problem (-1/2)∫1/(u²+1) du to (-1/2)∫ 1 dv, and that is -v/2.

Put back in what v was here, go back and collect all the integrals together,
and add a C to the collected terms, and there's the answer.