Practice decomposing rational functions and integrating the resulting pieces.
Evaluate \(\displaystyle\int\dfrac{1}{x(x+1)}\,dx\)
1. Multiply the entire equation by both factors
\(\dfrac{1}{x(x+1)} = \dfrac{A}{x} + \dfrac{B}{x+1}\)
Multiply both sides by \(x(x+1)\):
\(1 = A(x+1) + Bx\)
2. Select values of \(x\) to zero out a term
Let \(x = -1\): \(1 = A(0) + B(-1) \quad\Rightarrow\quad B = -1\)
Let \(x = 0\): \(1 = A(1) + B(0) \quad\Rightarrow\quad A = 1\)
3. Rewrite the integral
\(\displaystyle\int\dfrac{1}{x(x+1)}\,dx = \displaystyle\int\left(\dfrac{1}{x} – \dfrac{1}{x+1}\right)dx\)
4. Integrate term by term
\(\ln|x| – \ln|x+1| + C\)
5. Combine using log properties
\(\ln\left|\dfrac{x}{x+1}\right| + C\)
Distinct linear factors only
\(\displaystyle\int\dfrac{1}{x(x-2)}\,dx\)
1. Multiply by both factors
\(\dfrac{1}{x(x-2)} = \dfrac{A}{x} + \dfrac{B}{x-2}\)
\(1 = A(x-2) + Bx\)
2. Select values of \(x\)
Let \(x = 0\): \(1 = A(-2) \quad\Rightarrow\quad A = -\dfrac{1}{2}\)
Let \(x = 2\): \(1 = B(2) \quad\Rightarrow\quad B = \dfrac{1}{2}\)
3. Rewrite the integral
\(\displaystyle\int\dfrac{1}{x(x-2)}\,dx = \displaystyle\int\left(-\dfrac{1}{2x} + \dfrac{1}{2(x-2)}\right)dx\)
4. Integrate term by term
\(-\dfrac{1}{2}\ln|x| + \dfrac{1}{2}\ln|x-2| + C\)
5. Combine using log properties
\(\dfrac{1}{2}\ln\left|\dfrac{x-2}{x}\right| + C\)
\(\displaystyle\int\dfrac{3}{(x+1)(x-1)}\,dx\)
1. Multiply by both factors
\(\dfrac{3}{(x+1)(x-1)} = \dfrac{A}{x+1} + \dfrac{B}{x-1}\)
\(3 = A(x-1) + B(x+1)\)
2. Select values of \(x\)
Let \(x = -1\): \(3 = A(-2) \quad\Rightarrow\quad A = -\dfrac{3}{2}\)
Let \(x = 1\): \(3 = B(2) \quad\Rightarrow\quad B = \dfrac{3}{2}\)
3. Rewrite the integral
\(\displaystyle\int\dfrac{3}{(x+1)(x-1)}\,dx = \displaystyle\int\left(-\dfrac{3}{2(x+1)} + \dfrac{3}{2(x-1)}\right)dx\)
4. Integrate term by term
\(-\dfrac{3}{2}\ln|x+1| + \dfrac{3}{2}\ln|x-1| + C\)
5. Combine using log properties
\(\dfrac{3}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C\)
\(\displaystyle\int\dfrac{x+5}{x(x+1)}\,dx\)
1. Multiply by both factors
\(\dfrac{x+5}{x(x+1)} = \dfrac{A}{x} + \dfrac{B}{x+1}\)
\(x+5 = A(x+1) + Bx\)
2. Select values of \(x\)
Let \(x = 0\): \(5 = A(1) \quad\Rightarrow\quad A = 5\)
Let \(x = -1\): \(4 = B(-1) \quad\Rightarrow\quad B = -4\)
3. Rewrite the integral
\(\displaystyle\int\dfrac{x+5}{x(x+1)}\,dx = \displaystyle\int\left(\dfrac{5}{x} – \dfrac{4}{x+1}\right)dx\)
4. Integrate term by term
\(5\ln|x| – 4\ln|x+1| + C\)
Includes irreducible quadratic factors or difference of cubes
\(\displaystyle\int\dfrac{2x+3}{(x-1)(x+2)}\,dx\)
1. Multiply by both factors
\(\dfrac{2x+3}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}\)
\(2x+3 = A(x+2) + B(x-1)\)
2. Select values of \(x\)
Let \(x = 1\): \(5 = A(3) \quad\Rightarrow\quad A = \dfrac{5}{3}\)
Let \(x = -2\): \(-1 = B(-3) \quad\Rightarrow\quad B = \dfrac{1}{3}\)
3. Rewrite the integral
\(\displaystyle\int\dfrac{2x+3}{(x-1)(x+2)}\,dx = \displaystyle\int\left(\dfrac{5}{3(x-1)} + \dfrac{1}{3(x+2)}\right)dx\)
4. Integrate term by term
\(\dfrac{5}{3}\ln|x-1| + \dfrac{1}{3}\ln|x+2| + C\)
\(\displaystyle\int\dfrac{x+5}{(x-1)(x^{2}+4)}\,dx\)
1. Set up the decomposition
\(\dfrac{x+5}{(x-1)(x^{2}+4)} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^{2}+4}\)
2. Multiply by both factors
\(x+5 = A(x^{2}+4) + (Bx+C)(x-1)\)
3. Select a convenient value of \(x\)
Let \(x = 1\): \(6 = A(5) \quad\Rightarrow\quad A = \dfrac{6}{5}\)
4. Expand and match coefficients
\(A(x^{2}+4) + (Bx+C)(x-1) = (A+B)x^{2} + (-B+C)x + (4A-C)\)
This equals \(0\cdot x^{2} + 1\cdot x + 5\), so:
\(A + B = 0\)
\(-B + C = 1\)
\(4A – C = 5\)
With \(A = \dfrac{6}{5}\):
\(B = -\dfrac{6}{5}\), \(C = -\dfrac{1}{5}\)
5. Rewrite the integral
\(\displaystyle\int\left(\dfrac{6/5}{x-1} + \dfrac{-\frac{6}{5}x – \frac{1}{5}}{x^{2}+4}\right)dx\)
\(= \dfrac{6}{5}\displaystyle\int\dfrac{1}{x-1}\,dx – \dfrac{1}{5}\displaystyle\int\dfrac{6x+1}{x^{2}+4}\,dx\)
6. Split and integrate the quadratic piece
\(-\dfrac{1}{5}\displaystyle\int\dfrac{6x}{x^{2}+4}\,dx – \dfrac{1}{5}\displaystyle\int\dfrac{1}{x^{2}+4}\,dx\)
First integral (log form):
\(-\dfrac{1}{5}\cdot 3\ln(x^{2}+4) = -\dfrac{3}{5}\ln(x^{2}+4)\)
Second integral (arctan form):
\(-\dfrac{1}{5}\cdot\dfrac{1}{2}\arctan\dfrac{x}{2} = -\dfrac{1}{10}\arctan\dfrac{x}{2}\)
7. Final answer
\(\dfrac{6}{5}\ln|x-1| – \dfrac{3}{5}\ln(x^{2}+4) – \dfrac{1}{10}\arctan\dfrac{x}{2} + C\)
Ready to master partial fractions?