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Use partial fractions to solve the following integrals. A) \int \frac{3x^2-1}{x^3-x} dx B) \int...

Question:

Use partial fractions to solve the following integrals.

A) {eq}\int \frac{3x^2-1}{x^3-x} dx {/eq}

B) {eq}\int \frac{x^3+2x^2+2}{(x^2+1)^2} dx {/eq}

Partial Fraction Decomposition Method:

The integrals can solve it by several techniques. Commonly, you apply various techniques to solve an integral. That is, you can initially use integration by substitution and then apply partial fraction. The Partial Fraction Decomposition Method consists of decomposes the fraction given in several equivalent fractions. These fractions content various unknown constants. If you find the constant values, you obtain a new integral. The integral found can solve it with other easier methods.

Answer and Explanation:

{eq}\eqalign{ & {\text{We use the Partial Fraction Decomposition Method to solve the integral:}} \cr & {\text{A) }}\int {\frac{{3{x^2} - 1}}{{{x^3} - x}}} dx \cr & {\text{We factor in the bottom:}} \cr & \int {\frac{{3{x^2} - 1}}{{{x^3} - x}}} dx = \int {\frac{{3{x^2} - 1}}{{x\left( {{x^2} - 1} \right)}}} dx \cr & {\text{We have two factors}}{\text{. A linear factor }}x{\text{ and a quadratic factor }}{x^2} - 1.{\text{ We rewrite the integral in}} \cr & {\text{partial fractions:}} \cr & \int {\frac{{3{x^2} - 1}}{{{x^3} - x}}} dx = \int {\frac{{3{x^2} - 1}}{{x\left( {{x^2} - 1} \right)}}} dx = \int {\left( {\frac{A}{x} + \frac{{Bx + C}}{{{x^2} - 1}}} \right)} dx \cr & {\text{We solve:}} \cr & \frac{{3{x^2} - 1}}{{x\left( {{x^2} - 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} - 1}} \cr & {\text{We multiply by the bottom:}} \cr & 3{x^2} - 1 = A\left( {{x^2} - 1} \right) + \left( {Bx + C} \right)x \cr & 3{x^2} - 1 = A{x^2} - A + B{x^2} + Cx \cr & 3{x^2} - 1 = \left( {A + B} \right){x^2} + Cx - A \cr & {\text{We equal the coefficients of both polynomials}}{\text{. We find three equations:}} \cr & A + B = 3\,\,\, \Rightarrow \,\,{\text{Eq}}{\text{.1}} \cr & C = 0\,\,\, \Rightarrow \,\,{\text{Eq}}{\text{.2}} \cr & - A = - 1\,\,\, \Rightarrow \,\,{\text{Eq}}{\text{.3}} \cr & {\text{We solve:}} \cr & A = 1 \cr & {\text{We substitute }}A = 1{\text{ on the Eq}}{\text{.1:}} \cr & 1 + B = 3 \cr & B = 3 - 1 \cr & B = 2 \cr & {\text{We substitute this constants on the integral and solve:}} \cr & \int {\frac{{3{x^2} - 1}}{{{x^3} - x}}} dx = \int {\frac{{3{x^2} - 1}}{{x\left( {{x^2} - 1} \right)}}} dx = \int {\left( {\frac{A}{x} + \frac{{Bx + C}}{{{x^2} - 1}}} \right)} dx \cr & = \int {\left( {\frac{1}{x} + \frac{{2x + 0}}{{{x^2} - 1}}} \right)} dx = \int {\frac{{dx}}{x}} + 2\int {\frac{x}{{{x^2} - 1}}} dx \cr & {\text{We have two integrals}}{\text{. We solve the second integral using Integration by Substitution:}} \cr & u = {x^2} - 1 \cr & du = 2xdx \cr & \frac{1}{2}du = xdx \cr & = \int {\frac{{dx}}{x}} + 2*\frac{1}{2}\int {\frac{{du}}{u}} = \ln x + \ln u + K = \ln x + \ln \left( {{x^2} - 1} \right) + K = \ln \left[ {x\left( {{x^2} - 1} \right)} \right] + K \cr} {/eq}

{eq}\eqalign{ & {\text{We use the Partial Fraction Decomposition Method to solve the integral:}} \cr & {\text{B) }}\int {\frac{{{x^3} + 2{x^2} + 2}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr & {\text{We have a quadratic factor repeated }}{\left( {{x^2} + 1} \right)^2}{\text{. We rewrite the integral in}} \cr & {\text{partial fractions:}} \cr & \int {\frac{{{x^3} + 2{x^2} + 2}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx = \int {\left( {\frac{{Ax + B}}{{{x^2} + 1}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)} dx \cr & {\text{We solve:}} \cr & \frac{{{x^3} + 2{x^2} + 2}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{Ax + B}}{{{x^2} + 1}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & {\text{We multiply the fractions by }}{\left( {{x^2} + 1} \right)^2}{\text{:}} \cr & {x^3} + 2{x^2} + 2 = \left( {Ax + B} \right)\left( {{x^2} + 1} \right) + Cx + D \cr & {x^3} + 2{x^2} + 2 = A{x^3} + Ax + B{x^2} + B + Cx + D \cr & {x^3} + 2{x^2} + 2 = A{x^3} + B{x^2} + \left( {A + C} \right)x + \left( {B + D} \right) \cr & {\text{We equal the coefficients of both polynomials}}{\text{. We find four equations:}} \cr & A = 1\,\, \Rightarrow \,\,{\text{Eq}}{\text{.1}} \cr & B = 2\,\, \Rightarrow \,\,{\text{Eq}}{\text{.2}} \cr & A + C = 0\,\, \Rightarrow \,\,{\text{Eq}}{\text{.3}} \cr & B + D = 2\,\, \Rightarrow \,\,{\text{Eq}}{\text{.4}} \cr & {\text{We substitute Eq}}{\text{.1 on the Eq}}{\text{.3 and obtain }}C{\text{:}} \cr & 1 + C = 0 \cr & C = - 1 \cr & {\text{We substitute Eq}}{\text{.2 on the Eq}}{\text{.4 and obtain }}D{\text{:}} \cr & 2 + D = 2 \cr & D = 2 - 2 \cr & D = 0 \cr & {\text{We substitute this constants on the integral and solve:}} \cr & \int {\frac{{{x^3} + 2{x^2} + 2}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx = \int {\left( {\frac{{Ax + B}}{{{x^2} + 1}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)} dx = \int {\left( {\frac{{x + 2}}{{{x^2} + 1}} + \frac{{ - x + 0}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)} dx \cr & = \int {\left( {\frac{{x + 2}}{{{x^2} + 1}} - \frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)} dx = \int {\frac{x}{{{x^2} + 1}}dx} + 2\int {\frac{{dx}}{{{x^2} + 1}}} - \int {\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} \cr & {\text{We have three integrals}}{\text{. We solve the first and third integral using Integration by Substitution:}} \cr & u = {x^2} + 1 \cr & du = 2xdx \cr & \frac{1}{2}du = xdx \cr & = \frac{1}{2}\int {\frac{{du}}{u}} + 2\int {\frac{{dx}}{{{x^2} + 1}}} - \frac{1}{2}\int {\frac{{du}}{{{u^2}}}} = \frac{1}{2}\int {\frac{{du}}{u}} + 2\int {\frac{{dx}}{{{x^2} + 1}}} - \frac{1}{2}\int {{u^{ - 2}}du} = \frac{1}{2}\ln u + 2\arctan x + \frac{1}{2}{u^{ - 1}} + K \cr & = \frac{1}{2}\ln u + 2\arctan x + \frac{1}{{2u}} + K = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + 2\arctan x + \frac{1}{{2\left( {{x^2} + 1} \right)}} + K \cr} {/eq}


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Partial Fraction Decomposition: Rules & Examples

from High School Algebra I: Help and Review

Chapter 3 / Lesson 25
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