← Mathematics Course contents Section PDF ↗ Partial Fractions — Question 1 Question 1
Find
A A
and
B B ,
then evaluate the integral:
7 x + 1 ( x − 1 ) ( x + 2 ) = A x − 1 + B x + 2 . \frac{7x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}.
∫ 7 x + 1 ( x − 1 ) ( x + 2 ) d x . \int\frac{7x+1}{(x-1)(x+2)}\,dx.
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Solution
Step 1: Set up the decomposition.
7 x + 1 ( x − 1 ) ( x + 2 ) = A x − 1 + B x + 2 . \frac{7x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}.
Step 2: Clear the denominators. Multiply by
( x − 1 ) ( x + 2 ) (x-1)(x+2) :
7 x + 1 = A ( x + 2 ) + B ( x − 1 ) . 7x+1=A(x+2)+B(x-1).
Step 3: Find the coefficients.
x = 1 : 8 = 3 A ⇒ A = 8 3 , x = − 2 : − 13 = − 3 B ⇒ B = 13 3 . \begin{align*}
x=1:&\quad 8=3A \quad\Longrightarrow\quad A=\frac83,\\
x=-2:&\quad -13=-3B \quad\Longrightarrow\quad B=\frac{13}{3}.
\end{align*} Step 4: Substitute
and integrate each term.
∫ 7 x + 1 ( x − 1 ) ( x + 2 ) d x = ∫ ( 8 / 3 x − 1 + 13 / 3 x + 2 ) d x = 8 3 ∫ d x x − 1 + 13 3 ∫ d x x + 2 = 8 3 ln | x − 1 | + 13 3 ln | x + 2 | + C . \begin{align*}
\int\frac{7x+1}{(x-1)(x+2)}\,dx
&=\int\left(\frac{8/3}{x-1}+\frac{13/3}{x+2}\right)dx\\
&=\frac83\int\frac{dx}{x-1}+\frac{13}{3}\int\frac{dx}{x+2}\\
&=\boxed{\frac83\ln|x-1|+\frac{13}{3}\ln|x+2|+C}.
\end{align*}
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