Apostol (5.11.22): Finding $a,b$ s.t. $\int_0^1 (ax+b)(x^2+3x+2)^{-2} dx = \frac{3}{2}$

Am stuck on another question from Apostol "Calculus" Volume 1 (Section 5.11, Question 22). The question reads:

Determine a pair of numbers $a$ and $b$ such that $$\int_0^1 (ax+b)(x^2+3x+2)^{-2} dx = \frac{3}{2}$$

I observed immediately that $x^2+3x+2 = (x+2)(x+1)$ so we have:

$$\int_0^1 (ax+b)(x^2+3x+2)^{-2}dx = \int_0^1 \frac{ax+b}{(x+1)^2(x+2)^2} dx$$ But it's at this point that I seem to run into some problems. I tried various combinations with integration by parts, for example:

1. set $dv = ax+b$, $u = 1/((x+1)^{-2}(x+2)^{-2})$
2. set $u = (ax+b)/(x+2)^{-2}$, $dv = (x+1)^{-2}$
3. set $u = ax+b$, $dv = 1/((x+1)^{-2}(x+2)^{-2})$

But all of these seemed to turn into extremely unwieldy integrals, which (to me) goes against the point of integration by parts.

I also tried integration by substitution but I ran into similar problems with combinations such as

1. set $w = x+1$, so $x = w - 1$ and $w+1$ = $x+2$
2. Similarly for $w = x+2$

At this point I felt my options were running a little thin. However, I am now wondering if this exercise could be an application of the (weighted) mean value theorem and the second mean value theorem for integrals, that is, for some $c\in [0,1]$ and using the second mean value theorem for integrals, we would have $$\int_0^1 \frac{ax+b}{(x+1)^2(x+2)^2} dx = \frac{1}{4}\int_0^c(ax+b)dx + \frac{1}{36}\int_c^1(ax+b)dx$$ which (to me) would be fine since both functions are continuous in $[0, 1]$. Obviously I would need to do something similar using the (weighted) mean value theorem for integrals, since I would be solving for multiple unknowns values.

Please note, I am only looking for some (reasonably broad) hints as to the right direction to take to solve this problem, for example, is using the mean value theorem correct? If so, can I pick an arbitrary number for $c$? Or have I missed the point entirely?

• Breaking up the integrand into partial fractions and integrating might work. – solstafir Apr 18 '14 at 1:01
• I was actually trying to avoid using integration by partial fractions since that is not introduced until Chapter 6 in the book. – emjay Apr 18 '14 at 1:05
• Ah, I see. Then I would look more into the mean value theorem idea, although I am not sure whether or not that will work. – solstafir Apr 18 '14 at 1:06
• I had another thought. Don't factor the denominator, and let $x^2+3x+2=u$. Now observe how $ax+b=\frac{a}{2}(2x+3)+b-\frac{3a}{2}$. – solstafir Apr 18 '14 at 1:11
• Actually, you will run into complications with this because the second integral will have a quartic in the denominator but only constants in the numerator. Sandeep Silwal suggested a similar idea which might work better. – solstafir Apr 18 '14 at 1:16

Hint: Power rule! Notice that $ax+b$ looks awefully close do the derivative of $x^2+3x+2$. (In fact just let $ax+b$ be the derivative and just scale the integral by a constant.