Exercise 7.5: Integration by Partial Fractions
- Ex. 7.1
- Ex. 7.2
- Ex. 7.2
- Ex. 7.3
- Ex. 7.4
- Ex. 7.5
- Ex. 7.6
- Ex. 7.7
- Ex. 7.8
- Ex. 7.9
- Ex. 7.10
- Ex. 7.11
- Miscellaneous Exercise
Solution:
Let \(\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}\)
\(\Rightarrow x = A(x+2) + B(x+1) \quad \dots(1)\)
Putting \(x = -1\) in (1):
\(-1 = A(1) \Rightarrow A = -1\)
Putting \(x = -2\) in (1):
\(-2 = B(-1) \Rightarrow B = 2\)
Therefore,
\(\int \frac{x}{(x+1)(x+2)} dx = \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) dx\)
\(= -\log|x+1| + 2\log|x+2| + C\)
\(= \log(x+2)^2 - \log|x+1| + C\)
Solution:
The expression can be factored as \(\frac{1}{(x-3)(x+3)}\).
Let \(\frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3}\)
\(\Rightarrow 1 = A(x+3) + B(x-3) \quad \dots(1)\)
Putting \(x = 3\) in (1):
\(1 = 6A \Rightarrow A = \frac{1}{6}\)
Putting \(x = -3\) in (1):
\(1 = -6B \Rightarrow B = -\frac{1}{6}\)
Therefore,
\(\int \frac{dx}{x^2 - 9} = \frac{1}{6} \int \frac{dx}{x-3} - \frac{1}{6} \int \frac{dx}{x+3}\)
\(= \frac{1}{6} \log|x-3| - \frac{1}{6} \log|x+3| + C\)
\(= \frac{1}{6} \log \left| \frac{x-3}{x+3} \right| + C\)
Solution:
Let \(\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\)
\(\Rightarrow 3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)\)
Putting \(x=1\): \(2 = A(-1)(-2) \Rightarrow 2A = 2 \Rightarrow A = 1\)
Putting \(x=2\): \(5 = B(1)(-1) \Rightarrow -B = 5 \Rightarrow B = -5\)
Putting \(x=3\): \(8 = C(2)(1) \Rightarrow 2C = 8 \Rightarrow C = 4\)
Thus,
\(\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx\)
\(= \log|x-1| - 5\log|x-2| + 4\log|x-3| + C\)
Solution:
Let \(\frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\)
\(\Rightarrow x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)\)
Putting \(x=1\): \(1 = 2A \Rightarrow A = \frac{1}{2}\)
Putting \(x=2\): \(2 = -B \Rightarrow B = -2\)
Putting \(x=3\): \(3 = 2C \Rightarrow C = \frac{3}{2}\)
Therefore,
\(\int \frac{x}{(x-1)(x-2)(x-3)} dx = \frac{1}{2}\log|x-1| - 2\log|x-2| + \frac{3}{2}\log|x-3| + C\)
Solution:
Denominator \(x^2+3x+2 = (x+1)(x+2)\).
Let \(\frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}\)
\(\Rightarrow 2x = A(x+2) + B(x+1)\)
Putting \(x=-1\): \(-2 = A \Rightarrow A = -2\)
Putting \(x=-2\): \(-4 = -B \Rightarrow B = 4\)
Thus,
\(\int \frac{2x}{x^2+3x+2} dx = -2\log|x+1| + 4\log|x+2| + C\)
Solution:
The integrand is \(\frac{1-x^2}{x-2x^2}\). Since the degree of the numerator equals the degree of the denominator, we divide first.
\(\frac{1-x^2}{-2x^2+x} = \frac{1}{2} + \frac{1 - \frac{x}{2}}{x(1-2x)} = \frac{1}{2} + \frac{2-x}{2x(1-2x)}\)
Let \(\frac{2-x}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x}\)
\(\Rightarrow 2-x = A(1-2x) + Bx\)
Putting \(x=0\): \(A=2\)
Putting \(x=\frac{1}{2}\): \(\frac{3}{2} = \frac{B}{2} \Rightarrow B=3\)
Thus, Integrand = \(\frac{1}{2} + \frac{1}{2} \left( \frac{2}{x} + \frac{3}{1-2x} \right)\)
\(\int \left( \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1-2x)} \right) dx\)
\(= \frac{x}{2} + \log|x| + \frac{3}{2} \cdot \frac{\log|1-2x|}{-2} + C\)
\(= \frac{x}{2} + \log|x| - \frac{3}{4}\log|1-2x| + C\)
Solution:
Let \(\frac{x}{(x^2+1)(x-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1}\)
\(\Rightarrow x = (Ax+B)(x-1) + C(x^2+1)\)
Putting \(x=1\): \(1 = 2C \Rightarrow C = \frac{1}{2}\)
Comparing coefficients of \(x^2\): \(A+C=0 \Rightarrow A = -\frac{1}{2}\)
Comparing constant terms: \(-B+C=0 \Rightarrow B = \frac{1}{2}\)
Integral = \(\int \left( \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2+1} + \frac{\frac{1}{2}}{x-1} \right) dx\)
\(= -\frac{1}{2}\int \frac{x}{x^2+1} dx + \frac{1}{2}\int \frac{1}{x^2+1} dx + \frac{1}{2}\int \frac{1}{x-1} dx\)
\(= -\frac{1}{4}\log(x^2+1) + \frac{1}{2}\tan^{-1}x + \frac{1}{2}\log|x-1| + C\)
Solution:
Let \(\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}\)
\(\Rightarrow x = A(x-1)(x+2) + B(x+2) + C(x-1)^2\)
Putting \(x=1\): \(1 = 3B \Rightarrow B = \frac{1}{3}\)
Putting \(x=-2\): \(-2 = 9C \Rightarrow C = -\frac{2}{9}\)
Comparing coeff of \(x^2\): \(A+C=0 \Rightarrow A = \frac{2}{9}\)
Integral = \(\frac{2}{9}\log|x-1| + \frac{1}{3} \int (x-1)^{-2} dx - \frac{2}{9}\log|x+2| + C\)
\(= \frac{2}{9}\log \left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + C\)
Solution:
\(x^3-x^2-x+1 = x^2(x-1) - 1(x-1) = (x^2-1)(x-1) = (x+1)(x-1)^2\)
Let \(\frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\)
\(\Rightarrow 3x+5 = A(x-1)^2 + B(x+1)(x-1) + C(x+1)\)
Putting \(x=1\): \(8 = 2C \Rightarrow C=4\)
Putting \(x=-1\): \(2 = 4A \Rightarrow A=\frac{1}{2}\)
Coeff of \(x^2\): \(A+B=0 \Rightarrow B=-\frac{1}{2}\)
Integral = \(\frac{1}{2}\log|x+1| - \frac{1}{2}\log|x-1| - \frac{4}{x-1} + C\)
\(= \frac{1}{2} \log \left| \frac{x+1}{x-1} \right| - \frac{4}{x-1} + C\)
Solution:
Factor: \((x-1)(x+1)(2x+3)\)
Let \(\frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3}\)
\(x=1 \Rightarrow -1 = A(2)(5) \Rightarrow A = -\frac{1}{10}\)
\(x=-1 \Rightarrow -5 = B(-2)(1) \Rightarrow B = \frac{5}{2}\)
\(x=-\frac{3}{2} \Rightarrow -6 = C(-\frac{5}{2})(-\frac{1}{2}) \Rightarrow C = -\frac{24}{5}\)
Integral = \(-\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{24}{5} \frac{\log|2x+3|}{2} + C\)
\(= \frac{5}{2}\log|x+1| - \frac{1}{10}\log|x-1| - \frac{12}{5}\log|2x+3| + C\)
Solution:
Factors: \((x+1)(x-2)(x+2)\). Let \(\frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2}\)
\(x=-1 \Rightarrow -5 = A(-3)(1) \Rightarrow A = \frac{5}{3}\)
\(x=2 \Rightarrow 10 = B(3)(4) \Rightarrow B = \frac{5}{6}\)
\(x=-2 \Rightarrow -10 = C(-1)(-4) \Rightarrow C = -\frac{5}{2}\)
Integral = \(\frac{5}{3}\log|x+1| + \frac{5}{6}\log|x-2| - \frac{5}{2}\log|x+2| + C\)
Solution:
Divide numerator by denominator: \(x^3+x+1 = x(x^2-1) + 2x+1\).
Integrand = \(x + \frac{2x+1}{x^2-1} = x + \frac{2x+1}{(x-1)(x+1)}\)
Let \(\frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\)
\(x=1 \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}\)
\(x=-1 \Rightarrow -1 = -2B \Rightarrow B = \frac{1}{2}\)
Integral = \(\frac{x^2}{2} + \frac{3}{2}\log|x-1| + \frac{1}{2}\log|x+1| + C\)
Solution:
Let \(\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2}\)
\(2 = A(1+x^2) + (Bx+C)(1-x)\)
\(x=1 \Rightarrow 2 = 2A \Rightarrow A=1\)
Coeff \(x^2\): \(A-B=0 \Rightarrow B=1\)
Constant: \(A+C=2 \Rightarrow C=1\)
Integral = \(\int \frac{1}{1-x} dx + \int \frac{x}{1+x^2} dx + \int \frac{1}{1+x^2} dx\)
\(= -\log|1-x| + \frac{1}{2}\log(1+x^2) + \tan^{-1}x + C\)
Solution:
Let \(\frac{3x-1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}\)
\(3x-1 = A(x+2) + B\)
\(A=3\), \(2A+B=-1 \Rightarrow 6+B=-1 \Rightarrow B=-7\)
Integral = \(3\log|x+2| + \frac{-7}{(x+2)} \cdot (-1) \text{ (Oops, integral of } u^{-2} \text{ is } -u^{-1})\)
\(= 3\log|x+2| + \frac{7}{x+2} + C\)
Solution:
\(\frac{1}{x^4-1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} \right]\)
\(= \frac{1}{2(x-1)(x+1)} - \frac{1}{2(x^2+1)}\)
\(= \frac{1}{2} \cdot \frac{1}{2} \left( \frac{1}{x-1} - \frac{1}{x+1} \right) - \frac{1}{2(x^2+1)}\)
\(= \frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)}\)
Integral = \(\frac{1}{4}\log|x-1| - \frac{1}{4}\log|x+1| - \frac{1}{2}\tan^{-1}x + C\)
\(= \frac{1}{4}\log \left| \frac{x-1}{x+1} \right| - \frac{1}{2}\tan^{-1}x + C\)
Solution:
Multiply numerator and denominator by \(x^{n-1}\): \(\int \frac{x^{n-1}}{x^n(x^n+1)} dx\).
Put \(x^n = t \Rightarrow nx^{n-1}dx = dt\).
Integral = \(\frac{1}{n} \int \frac{dt}{t(t+1)} = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt\)
\(= \frac{1}{n} \left( \log|t| - \log|t+1| \right) + C\)
\(= \frac{1}{n} \log \left| \frac{x^n}{x^n+1} \right| + C\)
Solution:
Put \(\sin x = t \Rightarrow \cos x dx = dt\).
Integrand becomes \(\frac{1}{(1-t)(2-t)}\).
Partial fractions: \(\frac{1}{(1-t)(2-t)} = \frac{1}{1-t} - \frac{1}{2-t}\) (Check: \(\frac{(2-t)-(1-t)}{(1-t)(2-t)} = \frac{1}{\dots}\))
Integral = \(\int \frac{dt}{1-t} - \int \frac{dt}{2-t}\)
\(= -\log|1-t| - (-\log|2-t|) + C\)
\(= \log|2-t| - \log|1-t| + C = \log \left| \frac{2-\sin x}{1-\sin x} \right| + C\)
Solution:
Let \(x^2 = y\) for partial fractions. \(\frac{(y+1)(y+2)}{(y+3)(y+4)} = \frac{y^2+3y+2}{y^2+7y+12} = 1 - \frac{4y+10}{(y+3)(y+4)}\).
\(\frac{4y+10}{(y+3)(y+4)} = \frac{A}{y+3} + \frac{B}{y+4}\)
\(y=-3 \Rightarrow -2 = A\). \(y=-4 \Rightarrow -6 = -B \Rightarrow B=6\).
Integrand = \(1 - \left( \frac{-2}{x^2+3} + \frac{6}{x^2+4} \right) = 1 + \frac{2}{x^2+3} - \frac{6}{x^2+4}\)
Integral = \(x + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} - 3\tan^{-1}\frac{x}{2} + C\)
Solution:
Put \(x^2 = t \Rightarrow 2x dx = dt\).
Integrand = \(\frac{1}{(t+1)(t+3)} = \frac{1}{2} \left( \frac{1}{t+1} - \frac{1}{t+3} \right)\).
Integral = \(\frac{1}{2} \left( \log|t+1| - \log|t+3| \right) + C\)
\(= \frac{1}{2} \log \left( \frac{x^2+1}{x^2+3} \right) + C\)
Solution:
Use result from Q16 with \(n=4\), but note sign change if using \(t=x^4\).
\(\int \frac{x^3 dx}{x^4(x^4-1)}\). Let \(x^4=t\). \(\frac{1}{4}\int \frac{dt}{t(t-1)}\).
\(\frac{1}{4} \int \left( \frac{1}{t-1} - \frac{1}{t} \right) dt = \frac{1}{4} \log \left| \frac{t-1}{t} \right| + C\)
\(= \frac{1}{4} \log \left| \frac{x^4-1}{x^4} \right| + C\)
Solution:
Put \(e^x = t \Rightarrow e^x dx = dt \Rightarrow dx = \frac{dt}{t}\).
Integral = \(\int \frac{dt}{t(t-1)}\). Same as Q20 partial fractions.
\(= \int \left( \frac{1}{t-1} - \frac{1}{t} \right) dt = \log|t-1| - \log|t| + C\)
\(= \log \left| \frac{e^x-1}{e^x} \right| + C = \log|1 - e^{-x}| + C\)
Solution:
\(\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}\).
Integral = \(-\log|x-1| + 2\log|x-2| + C = \log \left| \frac{(x-2)^2}{x-1} \right| + C\).
Correct Answer: B
Solution:
\(\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}\).
Integral = \(\log|x| - \frac{1}{2}\log(x^2+1) + C\).
Correct Answer: A