Q.Prove that the definite integral from 1 to 3 of dx / [x^2 (x + 1)] = 2/3 + log(2/3).

Partial fractions give $A = -1,\; B = 2$, so the claim $A = 1,\, B = 2$ is False.

Decompose the integrand:

$\frac{x}{(1 + x)(2 + x)} = \frac{A}{1 + x} + \frac{B}{2 + x}.$

Multiply through by $(1+x)(2+x)$:

$x = A(2 + x) + B(1 + x).$

Put $x = -1$: $\;-1 = A(2 - 1) + B(0) = A \Rightarrow A = -1.$

…

Partial fractions give $\dfrac{1}{(x-8)(x+7)} = \dfrac{1}{15}\!\left(\dfrac{1}{x-8} - \dfrac{1}{x+7}\right)$, which integrates to $\dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c$.

Write

$\frac{1}{(x-8)(x+7)} = \frac{A}{x-8} + \frac{B}{x+7} \;\Rightarrow\; A(x+7) + B(x-8) = 1.$

Put $x = 8$: $\;A(15) = 1 \Rightarrow A = \dfrac{1}{15}$.

Put $x = -7$: $\;B(-15) = 1 \Rightarrow B = -\dfrac{1}{15}$.

Hence …

Partial fractions give $\dfrac{x}{(x+2)(x+3)} = \dfrac{-2}{x+2} + \dfrac{3}{x+3}$, so the blank preceding $\displaystyle\int \dfrac{3}{x+3}\,dx$ is $\displaystyle\int \dfrac{-2}{x+2}\,dx$.

Resolve the integrand into partial fractions:

$\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}.$

Clearing denominators: $x = A(x+3) + B(x+2)$.

• Put $x = -2$: $-2 = A(1) \Rightarrow A = -2$.
• Put $x = -3$: $-3 = B(-1) \Rightarrow B = 3$.

Hence

$\frac{x}{(x+2)(x+3)} = \frac{-2}{x+2} + \frac{3}{x+3},$

so …