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

Evaluate the value of the integral, ∫x2+x+1(x+1)2(x+2)\int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}}

Explanation

Solution

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=∫x2+x+1(x+1)2(x+2)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 x2+x+1(x+1)2(x+2)\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} as:
x2+x+1(x+1)2(x+2)=A(x+1)+B(x+1)2+C(x+2) =A(x+1)(x+2)+B(x+2)+C(x+1)2(x+1)2(x+2) =A(x2+3x+2)+Bx+2B+C(x2+2x+1)(x+1)2(x+2) =x2(A+C)+x(3A+B+2C)+(2A+2B+C)(x+1)2(x+2)  \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
⇒A+C=1−−−−(i) ⇒3A+B+2C=1−−−−(ii) ⇒2A+2B+C=1−−−−(iii)  \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)A = 1 - C - - - - (iv)
Substituting the value from the equation (iv) in the equation (ii) and (iii), we can write
⇒3A+B+2C=1 ⇒3(1−C)+B+2C=1 ⇒3−3C+B+2C=1 ⇒B−C=−2−−−−(v)  \Rightarrow 3A + B + 2C = 1 \\\ \Rightarrow 3(1 - C) + B + 2C = 1 \\\ \Rightarrow 3 - 3C + B + 2C = 1 \\\ \Rightarrow B - C = - 2 - - - - (v) \\\
Similarly,
⇒2A+2B+C=1 ⇒2(1−C)+2B+C=1 ⇒2−2C+2B+C=1 ⇒2B−C=−1−−−−(vi)  \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
⇒B−C=−2 ⇒B=−2+C−−−−(vii)  \Rightarrow B - C = - 2 \\\ \Rightarrow B = - 2 + C - - - - (vii) \\\
Substituting the value of the equation (vii) in the equation (vi) we get
⇒2B−C=−1 ⇒2(−2+C)−C=−1 ⇒−4+2C−C=−1 ⇒C=3  \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
⇒B=−2+C =−2+3 =1  \Rightarrow B = - 2 + C \\\ = - 2 + 3 \\\ = 1 \\\
Again, substituting the value of C as 3 in the equation (iv), we get
⇒A=1−C =1−3 =−2  \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:

⇒∫x2+x+1(x+1)2(x+2)dx=∫(A(x+1)+B(x+1)2+C(x+2))dx =∫(−2(x+1)+1(x+1)2+3(x+2))dx \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 ∫(A+B+C)dx=∫Adx+∫Bdx+∫Cdx\int {\left( {A + B + C} \right)dx} = \int {Adx} + \int {Bdx} + \int {Cdx} in the above function.

⇒∫x2+x+1(x+1)2(x+2)dx=∫(−2(x+1)+1(x+1)2+3(x+2))dx =∫−2(x+1)dx+∫1(x+1)2dx+∫3(x+2)dx \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 ∫dxx=ln⁡x\int {\dfrac{{dx}}{x}} = \ln x in the above equation.

⇒∫x2+x+1(x+1)2(x+2)dx=∫−2(x+1)dx+∫1(x+1)2dx+∫3(x+2)dx =−2ln⁡∣x+1∣+∫(x+1)−2dx+3ln⁡∣x+2∣+c =−2ln⁡∣x+1∣+3ln⁡∣x+2∣−1(x+1)+c \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 ∫x2+x+1(x+1)2(x+2)\int {\dfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} is −2ln⁡∣x+1∣+3ln⁡∣x+2∣−1(x+1)+c - 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 1(x+1)2=A(x+1)+B(x+1)2\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 1(x+2)(x2+4x+2)=Ax+Bx2+4x+2+Cx+2\dfrac{1}{{\left( {x + 2} \right)\left( {{x^2} + 4x + 2} \right)}} = \dfrac{{Ax + B}}{{{x^2} + 4x + 2}} + \dfrac{C}{{x + 2}} .