Partial Fractions in Integration - IB Mathematics Analysis and Approaches

Using the substitution $u = e^x$ , find $\int \frac{e^x}{e^{2x} - 3e^x + 2} dx$ .
Answer: $\ln|\frac{e^x-2}{e^x-1}| + C$ Let $u = e^x$ . Then $du = e^x dx$ and $e^{2x} = u^2$ . The integral transforms to $\int \frac{1}{u^2-3u+2} du = \int \frac{1}{(u-1)(u-2)} du$ . Decompose the new integrand: $\frac{1}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}$ . $1 = A(u-2) + B(u-1)$ . If $u=1$ , $1 = -A \Rightarrow A=-1$ . If $u=2$ , $1 = B$ . The integral in $u$ is $\int (\frac{-1}{u-1} + \frac{1}{u-2}) du = -\ln|u-1| + \ln|u-2| + C = \ln|\frac{u-2}{u-1}| + C$ . Substituting back $u = e^x$ gives $\ln|\frac{e^x-2}{e^x-1}| + C$ .
Evaluate $\int_{e^2}^{e^3} \frac{1}{x(\ln x)^2 - x \ln x} dx$ .
Answer: $\ln(4/3)$ First, simplify the integrand: $\frac{1}{x((\ln x)^2 - \ln x)} = \frac{1}{x\ln x(\ln x - 1)}$ . Let $u = \ln x$ . Then $du = \frac{1}{x} dx$ . The limits of integration also change: when $x=e^2$ , $u=\ln(e^2)=2$ . When $x=e^3$ , $u=\ln(e^3)=3$ . The integral becomes $\int_2^3 \frac{1}{u(u-1)} du$ . Decompose into partial fractions: $\frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1}$ . $1 = A(u-1) + Bu$ . If $u=0$ , $1 = -A \Rightarrow A=-1$ . If $u=1$ , $1 = B$ . The integral is $\int_2^3 (\frac{-1}{u} + \frac{1}{u-1}) du = [-\ln|u| + \ln|u-1|]_2^3 = [\ln|\frac{u-1}{u}|]_2^3$ . Evaluating at the limits: $\ln(\frac{3-1}{3}) - \ln(\frac{2-1}{2}) = \ln(\frac{2}{3}) - \ln(\frac{1}{2}) = \ln(\frac{2/3}{1/2}) = \ln(\frac{4}{3})$ .
The integration of $\frac{P(x)}{Q(x)}$ by partial fractions, where $P(x)$ and $Q(x)$ are polynomials, results in an antiderivative containing an $\arctan$ term. Which of the following must be true about the denominator $Q(x)$ ?
Answer: $Q(x)$ has an irreducible quadratic factor. The $\arctan$ function arises from integrals of the form $\int \frac{1}{u^2+a^2} du$ . In the context of partial fractions, this type of integral results from a term of the form $\frac{Bx+C}{x^2+px+q}$ , where the denominator is an irreducible quadratic factor (i.e., it has no real roots, $p^2-4q < 0$ ). After completing the square, the denominator can be written in the form $(x+k)^2+a^2$ , which leads to an $\arctan$ term upon integration. A repeated linear factor (A) leads to terms of the form $C(x-a)^{-n}$ and logarithmic terms. An improper fraction (C) requires long division first, but this does not guarantee an $\arctan$ term. Distinct linear factors (D) only produce logarithmic terms.
Find the indefinite integral $\int \frac{3x+5}{(x+1)^2} dx$ .
Answer: $3\ln|x+1| - \frac{2}{x+1} + C$ For a repeated linear factor, the decomposition is $\frac{3x+5}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}$ . Multiplying by $(x+1)^2$ gives $3x+5 = A(x+1) + B$ . Comparing coefficients of $x$ : $3=A$ . Comparing constant terms: $5=A+B \Rightarrow 5=3+B \Rightarrow B=2$ . The integral is $\int (\frac{3}{x+1} + \frac{2}{(x+1)^2}) dx$ . Integrating term by term: $\int \frac{3}{x+1} dx + \int 2(x+1)^{-2} dx = 3\ln|x+1| + 2\frac{(x+1)^{-1}}{-1} + C = 3\ln|x+1| - \frac{2}{x+1} + C$ .
Evaluate the definite integral $\int_3^4 \frac{2x}{x^2-x-2} dx$ .
Answer: $\frac{2}{3}\ln(5)$ First, factor the denominator: $x^2-x-2 = (x-2)(x+1)$ . Decompose the fraction: $\frac{2x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$ . This gives $2x = A(x+1) + B(x-2)$ . If $x=2$ , $4 = 3A \Rightarrow A = 4/3$ . If $x=-1$ , $-2 = -3B \Rightarrow B = 2/3$ . So we need to evaluate $\int_3^4 (\frac{4/3}{x-2} + \frac{2/3}{x+1}) dx$ . The antiderivative is $[\frac{4}{3}\ln|x-2| + \frac{2}{3}\ln|x+1|]_3^4$ . Evaluating at the limits: At $x=4$ : $\frac{4}{3}\ln(2) + \frac{2}{3}\ln(5)$ . At $x=3$ : $\frac{4}{3}\ln(1) + \frac{2}{3}\ln(4) = 0 + \frac{2}{3}\ln(2^2) = \frac{4}{3}\ln(2)$ . Subtracting the lower limit from the upper limit: $(\frac{4}{3}\ln(2) + \frac{2}{3}\ln(5)) - (\frac{4}{3}\ln(2)) = \frac{2}{3}\ln(5)$ .
Find the value of $k$ such that $\int_2^3 \frac{k}{x^2-1} dx = \ln(\frac{3}{2})$ .
Answer: 2 First, find the integral of $\frac{1}{x^2-1}$ . Decompose $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ . $1 = A(x+1) + B(x-1)$ . If $x=1$ , $1=2A \Rightarrow A=1/2$ . If $x=-1$ , $1=-2B \Rightarrow B=-1/2$ . So $\int \frac{1}{x^2-1} dx = \frac{1}{2}(\ln|x-1| - \ln|x+1|) = \frac{1}{2}\ln|\frac{x-1}{x+1}|$ . Evaluating the definite integral: $\int_2^3 \frac{k}{x^2-1} dx = \frac{k}{2} [\ln|\frac{x-1}{x+1}|]_2^3$ . $= \frac{k}{2} [\ln(\frac{2}{4}) - \ln(\frac{1}{3})] = \frac{k}{2} [\ln(\frac{1}{2}) + \ln(3)] = \frac{k}{2} \ln(\frac{3}{2})$ . Setting this equal to $\ln(\frac{3}{2})$ : $\frac{k}{2} \ln(\frac{3}{2}) = \ln(\frac{3}{2})$ . Therefore $\frac{k}{2} = 1$ , so $k = 2$ .
The integral $\int \frac{2x^2-3x+2}{x(x-1)^2} dx$ is equal to $A\ln|x| + B\ln|x-1| + \frac{C}{x-1} + K$ . What is the value of $A+B$ ?
Answer: 2 The partial fraction decomposition is $\frac{2x^2-3x+2}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$ . Multiply by the denominator: $2x^2-3x+2 = A(x-1)^2 + Bx(x-1) + Cx$ . To find A, let $x=0$ : $2 = A(-1)^2 \Rightarrow A=2$ . To find C, let $x=1$ : $2(1)^2 - 3(1) + 2 = C(1) \Rightarrow 1 = C$ . To find B, we can compare the coefficients of the $x^2$ term on both sides. On the left, the coefficient is 2. On the right, it is $A+B$ . So, $A+B = 2$ . Since we are asked for $A+B$ , the value is 2. We don't even need to find B explicitly, but if we did: $2+B=2 \Rightarrow B=0$ .
Which of the following integrals requires polynomial long division before applying partial fraction decomposition?
Answer: $\int \frac{x^3+1}{x^3-1} dx$ Polynomial long division is required when the degree of the numerator is greater than or equal to the degree of the denominator (i.e., the rational function is improper). A: Degree of numerator (2) < degree of denominator (3). Not required. B: Degree of numerator (3) is equal to degree of denominator (3). Required. C: Degree of numerator (1) < degree of denominator (3). Not required. D: Degree of numerator (0) < degree of denominator (3). Not required. Therefore, only the integral in option B requires long division before proceeding with partial fractions.
Which of the following is the antiderivative of $f(x) = \frac{x^4}{x^2+1}$ ?
Answer: $\frac{1}{3}x^3 - x + \arctan(x) + C$ The degree of the numerator is greater than the degree of the denominator, so we perform polynomial long division. $x^4$ divided by $x^2+1$ can be written as $\frac{x^4-1+1}{x^2+1} = \frac{(x^2-1)(x^2+1)+1}{x^2+1} = x^2-1 + \frac{1}{x^2+1}$ . Now we integrate this expression: $\int (x^2-1 + \frac{1}{x^2+1}) dx = \int x^2 dx - \int 1 dx + \int \frac{1}{x^2+1} dx$ . This evaluates to $\frac{x^3}{3} - x + \arctan(x) + C$ .
A suitable substitution for evaluating $\int \frac{\cos(x)}{\sin^2(x) - 4} dx$ leads to an integral in terms of $u$ . What is the result of this integration in terms of $u$ ?
Answer: $\frac{1}{4}\ln|\frac{u-2}{u+2}| + C$ Let $u = \sin(x)$ , so $du = \cos(x) dx$ . The integral becomes $\int \frac{1}{u^2-4} du$ . This requires partial fractions: $\frac{1}{(u-2)(u+2)} = \frac{A}{u-2} + \frac{B}{u+2}$ . $1 = A(u+2) + B(u-2)$ . If $u=2$ , $1 = 4A \Rightarrow A=1/4$ . If $u=-2$ , $1 = -4B \Rightarrow B=-1/4$ . The integral is $\int (\frac{1/4}{u-2} - \frac{1/4}{u+2}) du = \frac{1}{4}(\ln|u-2| - \ln|u+2|) + C$ . Using logarithm properties, this simplifies to $\frac{1}{4}\ln|\frac{u-2}{u+2}| + C$ .
Find $\int \frac{2}{(x-1)(x^2+1)} dx$ .
Answer: $\ln|x-1| - \frac{1}{2}\ln(x^2+1) - \arctan(x) + C$ Decompose the integrand: $\frac{2}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$ . $2 = A(x^2+1) + (Bx+C)(x-1)$ . Let $x=1$ : $2 = A(1^2+1) \Rightarrow 2 = 2A \Rightarrow A=1$ . Substitute $A=1$ and expand: $2 = (x^2+1) + (Bx+C)(x-1) = x^2+1+Bx^2-Bx+Cx-C$ . $2 = (1+B)x^2 + (-B+C)x + (1-C)$ . Comparing coefficients of $x^2$ : $0 = 1+B \Rightarrow B=-1$ . Comparing constant terms: $2 = 1-C \Rightarrow C=-1$ . The integral is $\int (\frac{1}{x-1} + \frac{-x-1}{x^2+1}) dx = \int \frac{1}{x-1}dx - \int \frac{x}{x^2+1}dx - \int \frac{1}{x^2+1}dx$ . Integrating each part gives $\ln|x-1| - \frac{1}{2}\ln(x^2+1) - \arctan(x) + C$ .
Given that $\int \frac{ax+b}{x^2-3x-4} dx = 2\ln|x-4| + \ln|x+1| + C$ , find the value of $a$ .
Answer: 3 The result of the integration tells us about the partial fraction decomposition of the integrand. The integral is $\int (\frac{2}{x-4} + \frac{1}{x+1}) dx$ . To find the original integrand $\frac{ax+b}{x^2-3x-4}$ , we combine the fractions: $\frac{2}{x-4} + \frac{1}{x+1} = \frac{2(x+1) + 1(x-4)}{(x-4)(x+1)}$ . $= \frac{2x+2+x-4}{x^2-3x-4} = \frac{3x-2}{x^2-3x-4}$ . By comparing $\frac{3x-2}{x^2-3x-4}$ with $\frac{ax+b}{x^2-3x-4}$ , we can see that $a=3$ and $b=-2$ . The question asks for the value of $a$ , which is 3.
Which expression is equivalent to $\int \frac{x^3+2x^2}{x^2-1} dx$ ?
Answer: $\frac{1}{2}x^2+2x + \frac{3}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C$ The integrand is an improper rational function, so first perform polynomial long division. Dividing $x^3+2x^2$ by $x^2-1$ gives a quotient of $x+2$ and a remainder of $x+2$ . So, $\frac{x^3+2x^2}{x^2-1} = x+2 + \frac{x+2}{x^2-1}$ . Now, decompose the remainder term: $\frac{x+2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ . This gives $x+2 = A(x+1) + B(x-1)$ . If $x=1$ , $3 = 2A \Rightarrow A=3/2$ . If $x=-1$ , $1 = -2B \Rightarrow B=-1/2$ . The integral becomes $\int (x+2 + \frac{3/2}{x-1} - \frac{1/2}{x+1}) dx$ . Integrating gives $\frac{1}{2}x^2+2x + \frac{3}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C$ .
Evaluate $\int_1^2 \frac{1}{x^3+x} dx$ .
Answer: $\frac{1}{2}\ln(\frac{8}{5})$ First, factor the denominator: $x^3+x = x(x^2+1)$ . Decompose the fraction: $\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$ . $1 = A(x^2+1) + (Bx+C)x = (A+B)x^2+Cx+A$ . Comparing coefficients: $A=1$ , $C=0$ , $A+B=0 \Rightarrow B=-1$ . The integral is $\int_1^2 (\frac{1}{x} - \frac{x}{x^2+1}) dx$ . The antiderivative is $[\ln|x| - \frac{1}{2}\ln(x^2+1)]_1^2$ . Evaluate at the limits: At $x=2$ : $\ln(2) - \frac{1}{2}\ln(5)$ . At $x=1$ : $\ln(1) - \frac{1}{2}\ln(2) = -\frac{1}{2}\ln(2)$ . Subtracting: $(\ln(2) - \frac{1}{2}\ln(5)) - (-\frac{1}{2}\ln(2)) = \frac{3}{2}\ln(2) - \frac{1}{2}\ln(5)$ . Using logarithm properties: $\frac{1}{2}(3\ln(2) - \ln(5)) = \frac{1}{2}(\ln(2^3) - \ln(5)) = \frac{1}{2}\ln(\frac{8}{5})$ .
Find $\int \frac{5}{x(x^2+4)} dx$ .
Answer: $\frac{5}{4}\ln|x| - \frac{5}{8}\ln(x^2+4) + C$ The decomposition for the irreducible quadratic factor is $\frac{5}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4}$ . Multiplying by the denominator gives $5 = A(x^2+4) + (Bx+C)x$ . Expanding gives $5 = (A+B)x^2 + Cx + 4A$ . Comparing coefficients: Constant term: $5 = 4A \Rightarrow A=5/4$ . Coefficient of $x$ : $0 = C$ . Coefficient of $x^2$ : $0 = A+B \Rightarrow B = -A = -5/4$ . The integral is $\int (\frac{5/4}{x} - \frac{5/4 x}{x^2+4}) dx$ . This equals $\frac{5}{4}\ln|x| - \frac{5}{4}\int \frac{x}{x^2+4} dx$ . For the second integral, use substitution $u=x^2+4$ , $du=2x dx$ . The integral becomes $\int \frac{1}{2u} du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+4)$ . The final result is $\frac{5}{4}\ln|x| - \frac{5}{4}(\frac{1}{2}\ln(x^2+4)) + C = \frac{5}{4}\ln|x| - \frac{5}{8}\ln(x^2+4) + C$ .
Find the indefinite integral $\int \frac{4}{x^2-4} dx$ .
Answer: $\ln|\frac{x-2}{x+2}| + C$ First, decompose the integrand into partial fractions. The denominator factors as $x^2-4 = (x-2)(x+2)$ . Let $\frac{4}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$ . Multiplying by the denominator gives $4 = A(x+2) + B(x-2)$ . To find A, let $x=2$ : $4 = A(2+2) \Rightarrow 4 = 4A \Rightarrow A=1$ . To find B, let $x=-2$ : $4 = B(-2-2) \Rightarrow 4 = -4B \Rightarrow B=-1$ . So the integral becomes $\int (\frac{1}{x-2} - \frac{1}{x+2}) dx$ . Integrating term by term gives $\ln|x-2| - \ln|x+2| + C$ . Using the logarithm property $\ln a - \ln b = \ln(a/b)$ , the result is $\ln|\frac{x-2}{x+2}| + C$ .
The integral $\int \frac{2x^2+4}{x(x^2+4)} dx$ is equal to:
Answer: $\ln|x| + \frac{1}{2}\ln(x^2+4) + C$ Decompose the fraction: $\frac{2x^2+4}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4}$ . $2x^2+4 = A(x^2+4) + (Bx+C)x = (A+B)x^2+Cx+4A$ . Comparing coefficients: Constant term: $4=4A \Rightarrow A=1$ . Coefficient of $x$ : $C=0$ . Coefficient of $x^2$ : $2=A+B \Rightarrow 2=1+B \Rightarrow B=1$ . The integral becomes $\int (\frac{1}{x} + \frac{x}{x^2+4}) dx$ . $= \int \frac{1}{x} dx + \int \frac{x}{x^2+4} dx$ . This evaluates to $\ln|x| + \frac{1}{2}\ln(x^2+4) + C$ . The second part is a u-substitution where $u=x^2+4$ .

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