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Find constants to make a given integral equation true

by RoRi

Find constants a, b such that

    \[ \int_0^1 (ax+b)(x^2 + 3x + 2)^{-2} \, dx = \frac{3}{2}. \]

We would like to make the substitution,

    \[ u = x^2 + 3x + 2 \qquad \implies \qquad du = 2x + 3. \]

To do this we need the ax+b in the numerator of the integrand to be 2x+3. Let us assume there is some constant c we can multiply by such that c(ax+b) = 2x+3. Then we have

    \begin{align*} \frac{1}{c} \int_0^1 \frac{2x+3}{(x^2+3x+2)^2} \, dx &= \frac{1}{c} \int_{u(0)}^{u(1)} \frac{du}{u^2} \\ &= \frac{1}{c} \left. \left( -\frac{1}{u} \right) \right_{u(0)}^{u(1)} \\ &= \frac{1}{c} \left. \left( - \frac{1}{x^2+3x+2} \right) \right|_0^1 \\ &= \frac{1}{c} \left( \frac{1}{3} \right). \end{align*}

Since by assumption this integral equals \frac{3}{2} so we have

    \[ \frac{1}{3c} =\frac{3}{2} \qquad \implies \qquad c = \frac{2}{9}. \]

Putting this back into our original expression,

    \begin{align*} \frac{2}{9} (ax+b) = 2x + 3 && \implies && \frac{2}{9}a &= 2 \qquad \frac{2}{9} b = 3 \\ && \implies && a &= 9 \qquad b = \frac{27}{2}. \end{align*}

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