Evaluate the integral. {eq}\displaystyle \int_0^1 \dfrac {x^2 + x + 1} {(x + 1)^2 (x + 2)}\ dx {/eq}

Question:

Evaluate the integral.

{eq}\displaystyle \int_0^1 \dfrac {x^2 + x + 1} {(x + 1)^2 (x + 2)}\ dx {/eq}

Partial Fraction Decomposition:

Partial fraction decomposition/expansion is a way to expand a complex rational expression. In simple words, we break a complex rational expression into various parts. It provides various rules to break rational expression. The following integration formula helps us to solve the given definite integral problem.

\begin{align*} \int {{{\left( {x \pm a} \right)}^n}} &= \dfrac{{{{\left( {x \pm a} \right)}^{n + 1}}}}{{n + 1}} + C\\[0.3cm] \int {\dfrac{{dx}}{{x \pm a}}} &= \ln \left| {x \pm a} \right| + C\\[0.3cm] \int\limits_n^m {f\left( x \right)} \;dx &= F\left( m \right) - F\left( n \right) \end{align*}

Given Data:

• The given definite integral is {eq}\displaystyle \int\limits_0^1 {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} \;dx {/eq}.

Apply the partial fraction decomposition method to expand the integrand.

\begin{align*} \dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} &= \dfrac{A}{{x + 1}} + \dfrac{B}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{C}{{x + 2}}\\[0.3cm] \dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} &= \dfrac{{A\left( {x + 1} \right)\left( {x + 2} \right) + B\left( {x + 2} \right) + C{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}\\[0.3cm] {x^2} + x + 1 &= A\left( {x + 1} \right)\left( {x + 2} \right) + B\left( {x + 2} \right) + C{\left( {x + 1} \right)^2}\\[0.3cm] {x^2} + x + 1 &= A\left( {{x^2} + 3x + 2} \right) + B\left( {x + 2} \right) + C\left( {{x^2} + 2x + 1} \right)\\[0.3cm] {x^2} + x + 1 &= \left( {A + C} \right){x^2} + \left( {3A + B + 2C} \right)x + \left( {2A + 2B + C} \right) \end{align*}

Compare the coefficient of both sides.

\begin{align*} A + C &= 1 \Leftrightarrow A = 1 - C \Leftrightarrow A = - 2\\[0.3cm] 3A + B + 2C &= 1 \Leftrightarrow 3\left( {1 - C} \right) + B + 2C = 1 \Leftrightarrow - C + B = - 2 \Leftrightarrow B = - 2 + C \Leftrightarrow B = 1\\[0.3cm] 2A + 2B + C &= 1 \Leftrightarrow 2\left( {1 - C} \right) + 2\left( { - 2 + C} \right) + C = 1 \Leftrightarrow - 2 + C = 1 \Leftrightarrow C = 3 \end{align*}

So,

\begin{align*} \int\limits_0^1 {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} \;dx &= \int\limits_0^1 {\left( {\dfrac{{ - 2}}{{x + 1}} + \dfrac{1}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{3}{{\left( {x + 2} \right)}}} \right)} \;dx\\[0.3cm] &= \left[ { - 2\ln \left| {x + 1} \right| + \left( {\dfrac{{{{\left( {x + 1} \right)}^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) + 3\ln \left| {x + 2} \right|} \right]_0^1\\[0.3cm] &= \left[ { - 2\ln \left| {x + 1} \right| - \dfrac{1}{{x + 1}} + 3\ln \left| {x + 2} \right|} \right]_0^1\\[0.3cm] &= \left( { - 2\ln \left| 2 \right| - \dfrac{1}{2} + 3\ln \left| 3 \right|} \right) - \left( { - 2\ln \left| 1 \right| - 1 + 3\ln \left| {0 + 2} \right|} \right)\\[0.3cm] &= - 2\ln \left( 2 \right) - \dfrac{1}{2} + 3\ln \left( 3 \right) + 0 + 1 - 3\ln \left( 2 \right)\\[0.3cm] &= - 5\ln \left( 2 \right) + 3\ln \left( 3 \right) + \dfrac{1}{2} \end{align*}

 Thus, the solution of the given definite integral is {eq}- 5\ln \left( 2 \right) + 3\ln \left( 3 \right) + \dfrac{1}{2} {/eq}.