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Find the partial fractions decomposition and an antiderivative. \frac{2x + 3}{x^2 + 2x + 1}

Question:

Find the partial fractions decomposition and an antiderivative.

{eq}\frac{2x + 3}{x^2 + 2x + 1} {/eq}

Partial Fraction Decomposition:

We have to find the partial fractions decomposition and an antiderivative. Use the partial fractions decomposition, then substitute the value into antiderivative and use integration substitution method to find desired antiderivative.

Answer and Explanation:

Use partial fraction decomposition and we have $$\begin{align*} \frac{{2x + 3}}{{{{\left( {x + 1} \right)}^2}}} &= \frac{{{a_0}}}{{x + 1}} + \frac{{{a_1}}}{{{{\left( {x + 1} \right)}^2}}}.....\left( 1 \right)\\ &= \frac{{{a_0}\left( {x + 1} \right) + {a_1}}}{{{{\left( {x + 1} \right)}^2}}}\\ \frac{{2x + 3}}{{{{\left( {x + 1} \right)}^2}}} &= \frac{{{a_0}x + \left( {{a_0} + {a_1}} \right)}}{{{{\left( {x + 1} \right)}^2}}}. \end{align*} $$

Compare the both sides and we have $$\begin{align*} {a_0} &= 2\\ {a_0} + {a_1} &= 3 \Rightarrow {a_1} = 1. \end{align*} $$

Thus, the partial fraction decomposition will be {eq}\frac{{2x + 3}}{{{{\left( {x + 1} \right)}^2}}} = \frac{2}{{x + 1}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}. {/eq}

Substitute the value into the antiderivative and we have $$\begin{align*} I &= \int {\frac{{2x + 3}}{{{{\left( {x + 1} \right)}^2}}}} dx\\ &= \int {\left( {\frac{2}{{x + 1}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)dx} \\ I &= 2 \cdot \int {\frac{1}{{x + 1}}dx} + \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}dx} .....\left( 2 \right). \end{align*} $$

We have $$\begin{align*} \left( {let} \right)A &= \int {\frac{1}{{x + 1}}dx} .....\left( 3 \right)\\ \left( {let} \right)B &= \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}dx} .....\left( 4 \right). \end{align*} $$

Use integration substitution method into equation (3) and we have $$\begin{align*} u &= x + 1\\ \frac{{du}}{{dx}} &= \frac{d}{{dx}}\left( {x + 1} \right)\\ \frac{{du}}{{dx}} &= 1\\ du &= dx. \end{align*} $$

Thus, the integration will be $$\begin{align*} \left( {let} \right)A &= \int {\frac{1}{{x + 1}}dx} \\ &= \int {\frac{1}{u}du} \\ &= \ln \left| u \right| + {C_1}\\ A &= \ln \left| {x + 1} \right| + {C_1}. \end{align*} $$

Again, use integration substitution method into equation (3) and we have $$\begin{align*} u &= x + 1\\ \frac{{du}}{{dx}} &= \frac{d}{{dx}}\left( {x + 1} \right)\\ \frac{{du}}{{dx}} &= 1\\ du &= dx. \end{align*} $$

Thus, the integration will be $$\begin{align*} \left( {let} \right)B &= \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}dx} \\ &= \int {\frac{1}{{{u^2}}}du} \\ &= \int {{u^{ - 2}}du} \\ &= \frac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}} + {C_2}\\ &= - \frac{1}{u} + {C_2}\\ B &= - \frac{1}{{x + 1}} + {C_2}. \end{align*} $$

Substitute the values into equation (2) and we have $$\begin{align*} I &= 2 \cdot \int {\frac{1}{{x + 1}}dx} + \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}dx} \\ &= 2 \cdot \ln \left| {x + 1} \right| + {C_1} - \frac{1}{{x + 1}} + {C_2}\\ &= 2 \cdot \ln \left| {x + 1} \right| - \frac{1}{{x + 1}} + \left( {{C_1} + {C_2}} \right)\\ I &= 2 \cdot \ln \left| {x + 1} \right| - \frac{1}{{x + 1}} + C, \end{align*} $$ where "C" is an integral constant.


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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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