Question

Evaluate the value of the integral, $ \int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} $

Answer 1

Hint: In this question, we need to evaluate the given equation. For this, we will use the partial fraction method and apply the defined integral formulae.

Complete step-by-step answer:
Let the given integral be I so, we can write $ I = \int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} $ .
Now, by following the partial fraction method, we can write the function $ \dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} $ as:
 $
  \dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{{{(x + 1)}^2}}} + \dfrac{C}{{(x + 2)}} \\
   = \dfrac{{A(x + 1)(x + 2) + B(x + 2) + C{{(x + 1)}^2}}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} \\
   = \dfrac{{A({x^2} + 3x + 2) + Bx + 2B + C({x^2} + 2x + 1)}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} \\
   = \dfrac{{{x^2}(A + C) + x(3A + B + 2C) + (2A + 2B + C)}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} \\
$
Now, comparing the coefficients of both sides of the above equation, we get
 $
\Rightarrow A + C = 1 - - - - (i) \\
\Rightarrow 3A + B + 2C = 1 - - - - (ii) \\
\Rightarrow 2A + 2B + C = 1 - - - - (iii) \\
  $
Solving the above equations to determine the value of the constants A, B and C.
From the equation (i), we can write $ A = 1 - C - - - - (iv) $
Substituting the value from the equation (iv) in the equation (ii) and (iii), we can write
 $
\Rightarrow 3A + B + 2C = 1 \\
\Rightarrow 3(1 - C) + B + 2C = 1 \\
\Rightarrow 3 - 3C + B + 2C = 1 \\
\Rightarrow B - C = - 2 - - - - (v) \\
  $
Similarly,
 $
\Rightarrow 2A + 2B + C = 1 \\
\Rightarrow 2(1 - C) + 2B + C = 1 \\
\Rightarrow 2 - 2C + 2B + C = 1 \\
\Rightarrow 2B - C = - 1 - - - - (vi) \\
  $
Solving equations (v) and (vi).
From the equation (v), we get
 $
\Rightarrow B - C = - 2 \\
\Rightarrow B = - 2 + C - - - - (vii) \\
  $
Substituting the value of the equation (vii) in the equation (vi) we get
 $
\Rightarrow 2B - C = - 1 \\
\Rightarrow 2( - 2 + C) - C = - 1 \\
\Rightarrow - 4 + 2C - C = - 1 \\
\Rightarrow C = 3 \\
  $
Substituting the value of C as 3 in the equation (vii), we get
 $
\Rightarrow B = - 2 + C \\
   = - 2 + 3 \\
   = 1 \\
  $
Again, substituting the value of C as 3 in the equation (iv), we get
 $
\Rightarrow A = 1 - C \\
   = 1 - 3 \\
   = - 2 \\
  $
Hence, the values of the constant A, B and C are -2, 1 and 3 respectively.
So, the given integral function can be re-written as:
\[
\Rightarrow \int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}dx} = \int {\left( {\dfrac{A}{{(x + 1)}} + \dfrac{B}{{{{(x + 1)}^2}}} + \dfrac{C}{{(x + 2)}}} \right)} dx \\
   = \int {\left( {\dfrac{{ - 2}}{{(x + 1)}} + \dfrac{1}{{{{(x + 1)}^2}}} + \dfrac{3}{{(x + 2)}}} \right)} dx \\
 \]
Now, applying the property of the integration function $ \int {\left( {A + B + C} \right)dx} = \int {Adx} + \int {Bdx} + \int {Cdx} $ in the above function.
\[
\Rightarrow \int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}dx} = \int {\left( {\dfrac{{ - 2}}{{(x + 1)}} + \dfrac{1}{{{{(x + 1)}^2}}} + \dfrac{3}{{(x + 2)}}} \right)} dx \\
   = \int {\dfrac{{ - 2}}{{(x + 1)}}dx} + \int {\dfrac{1}{{{{(x + 1)}^2}}}dx} + \int {\dfrac{3}{{(x + 2)}}dx} \\
 \]
Again, applying the property of the integration function $ \int {\dfrac{{dx}}{x}} = \ln x $ in the above equation.
\[
\Rightarrow \int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}dx} = \int {\dfrac{{ - 2}}{{(x + 1)}}dx} + \int {\dfrac{1}{{{{(x + 1)}^2}}}dx} + \int {\dfrac{3}{{(x + 2)}}dx} \\
   = - 2\ln \left| {x + 1} \right| + \int {{{(x + 1)}^{ - 2}}dx} + 3\ln \left| {x + 2} \right| + c \\
   = - 2\ln \left| {x + 1} \right| + 3\ln \left| {x + 2} \right| - \dfrac{1}{{(x + 1)}} + c \\
 \]
Hence, we can see that the value of the integral $ \int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} $ is \[ - 2\ln \left| {x + 1} \right| + 3\ln \left| {x + 2} \right| - \dfrac{1}{{(x + 1)}} + c\] where, ‘c’ is the integral constant.

Note: It is interesting to note here that for the partial fraction method, if the denominator term is a raised to the power terms then, we need to bifurcate it as $ \dfrac{1}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{A}{{\left( {x + 1} \right)}} + \dfrac{B}{{{{\left( {x + 1} \right)}^2}}} $ where as if the denominator includes a quadratic equation then, we need to bifurcate it as $ \dfrac{1}{{\left( {x + 2} \right)\left( {{x^2} + 4x + 2} \right)}} = \dfrac{{Ax + B}}{{{x^2} + 4x + 2}} + \dfrac{C}{{x + 2}} $ .