Partial Fractions — Question 1

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Question 1

Find AA and BB, then evaluate the integral: 7x+1(x−1)(x+2)=Ax−1+Bx+2.\frac{7x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}. ∫7x+1(x−1)(x+2)dx.\int\frac{7x+1}{(x-1)(x+2)}\,dx.

Original worksheet page 1: question and worked solution for 1-4-001
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Question 1 – Solution

Step 1: Set up the decomposition. 7x+1(x−1)(x+2)=Ax−1+Bx+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): 7x+1=A(x+2)+B(x−1).7x+1=A(x+2)+B(x-1). Step 3: Find the coefficients. x=1:8=3A⇒A=83,x=−2:−13=−3B⇒B=133.\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. ∫7x+1(x−1)(x+2)dx=∫(8/3x−1+13/3x+2)dx=83∫dxx−1+133∫dxx+2=83ln⁡|x−1|+133ln⁡|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*}

Original worksheet page 2: question and worked solution for 1-4-001

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