# 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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from High School Algebra I: Help and Review

Chapter 3 / Lesson 25