Solving an equation involving an integral: $\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$

Question

Determine a pair of number $a$ and $b$ for which

$$\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$$

I tried putting $x$ as $1-x$ as the integral wouldn't change but could not move forward from there so can you please suggest me what should I do next.

Answer 1

Hint. One may observe that $$ (x^2+3x+2)=(x+1)(x+2) $$ leading to the following partial fraction decomposition $$ \begin{align} \frac{ax+b}{(x^2+3x+2)^2}&=\frac{ax+b}{(x+1)^2(x+2)^2} \\\\&=\frac{-a+b}{(x+1)^2}+\frac{3 a-2 b}{x+1}+\frac{-2 a+b}{(x+2)^2}+\frac{-3 a+2 b}{x+2}. \end{align} $$ Integrating each term gives $$ \int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx= \frac{-5a+4b}6+(3a-2b)(2\ln 2-\ln 3) $$ Can you take it from here?

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Edit. Partial fraction decomposition. $$ \begin{align} \frac{ax+b}{(x+1)^2(x+2)^2}&=\frac{A_1}{(x+1)^2}+\frac{B_1}{x+1}+\frac{A_2}{(x+2)^2}+\frac{B_2}{x+2}. \end{align} $$ By multiplying throughout by $(x+1)^2$ One has $$ \begin{align} \frac{ax+b}{(x+2)^2}&=A_1+(x+1)B_1+(x+1)^2\left(\frac{A_2}{(x+2)^2}+\frac{B_2}{x+2}\right). \end{align} $$ Putting $x=-1$ gives $A_1=-a+b$. Similarly by multiplying throughout by $(x+2)^2$ putting $x=-2$ gives $A_2=-2a+b$. Then by multiplying throughout by $(x+1)$ letting $x \to \infty$ gives $B_1+B_2=0$, finally $x=0$ gives $B_1=3a-2b=-B_2$. We are done.

Answer 2

Let $$I = \int_{0}^{1}\frac{ax+b}{(x^2+3x+2)^2}dx = \frac{a}{2}\int_{0}^{1}\underbrace{\frac{2x+3}{(x^2+3x+2)^2}dx}_{J}+\left(b-\frac{3a}{2}\right)\underbrace{\int_{0}^{1}\frac{1}{(x^2+3x+2)^2}dx}_{K}$$

So $$J = \int_{0}^{1}\frac{2x+3}{(x^2+3x+2)^2}dx = -\left[\frac{1}{x^2+3x+2}\right]_{0}^{1} = \left(\frac{1}{2}-\frac{1}{6}\right) = \frac{1}{3}$$

Similarly $$K = \int_{0}^{1}\frac{1}{(x^2+3x+2)^2}dx=\int_{0}^{1}\left[\frac{1}{x+1}-\frac{1}{x+2}\right]^2dx$$

So $$K = \int_{0}^{1}\frac{1}{(x+1)^2}dx+\int_{0}^{1}\frac{1}{(x+2)^2}dx-2\int_{0}^{1}\left[\frac{1}{x+1}-\frac{1}{x+2}\right]dx$$

So $$K = -\left[\frac{1}{x+1}\right]_{0}^{1}-\left[\frac{1}{x+2}\right]_{0}^{1}-2\left[\ln(x+1)-\ln(x+2)\right]_{0}^{1}$$

So $$K = \frac{1}{2}+\frac{1}{6}+2\left[\ln(4)-\ln(3)\right]=\frac{2}{3}+2\left[\ln\left|\frac{4}{3}\right|\right]$$

So $$I = \frac{a}{6}+\left(b-\frac{3a}{2}\right)\cdot \left(\frac{2}{3}+2\ln \left\|\frac{4}{3}\right\|\right) = \frac{5}{2}$$

Answer 3

This question, or one very similar to it, appeared on a diagnostic examination for incoming freshmen at the California Institute of Technology.

The first observation is that the system is underdetermined: for we may factor out $b$ and let $r = a/b$ to express the integrand as $$f(x) = b \cdot \frac{rx+1}{(x^2+3x+2)^2}.$$ For $r > 0$ and $x \in [0,1]$, we see that $f$ is obviously positive and bounded; thus there always exists a choice of $b$ for which the integral of $f$ on $[0,1]$ could be made equal to $5/2$. Thus, it suffices to choose a simple ratio for $a/b$, one that leads to a trivial antiderivative. To this end, what immediately comes to mind is $$r = \frac{a}{b} = \frac{2}{3},$$ so that we have $$k \int_{x=0}^1 \frac{2x+3}{(x^2 + 3x + 2)^2} \, dx = k \left[ - \frac{1}{x^2 + 3x + 2} \right]_{x=0}^1 = k \left(-\frac{1}{6} + \frac{1}{2} \right) = \frac{k}{3},$$ where $a = 2k$, $b = 3k$. Then the choice $k = \frac{15}{2}$ satisfies the desired relationship, or $a = 15$, $b = 45/2$.

The goal here is to not do any more work than is necessary to obtain a rigorous solution. It is not necessary to find a general antiderivative for $f$ in terms of $a, b$. By recognizing that the given condition does not possess a unique solution, and noting that we are only asked to find one instance of a solution, a lot of extra effort is saved.