Question

Evaluate the Integral: \(\int {\frac{{{x^3} + {x^2} + 2x + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}} dx\)

Short answer

The value of the given Integral can be given as:

\(\frac{1}{2}\ln |{x^2} + 1| + \frac{1}{{\sqrt 2 }}\ln |{x^2} + 1| + {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + c\)

Step-by-step solution

Apply Partial Function decomposition.

\(\begin{array}{l}\frac{{{x^3} + {x^2} + 2x + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}} = \frac{{Ax + B}}{{{x^2} + 1}} + \frac{{Cx + D}}{{{x^2} + 2}}\\\therefore {x^3} + {x^2} + 2x + 1 = \left( {Ax + B} \right)\left( {{x^2} + 2} \right) + \left( {Cx + D} \right)\left( {{x^2} + 1} \right)\\\therefore {x^3} + {x^2} + 2x + 1 = A{x^2} + 2Ax + B{x^2} + 2B + C{x^3} + Cx + D{x^2} + D\\\therefore {x^3} + {x^2} + 2x + 1 = \left( {A + C} \right){x^3} + \left( {B + D} \right){x^2} + \left( {2A + C} \right)x + 2B + D\end{array}\)

Compare thecoefficients of the above equation

\(\begin{array}{l}\therefore A + C = 1 \to \left( 1 \right)\\B + D = 1 \to \left( 2 \right)\\2A + C = 2 \to \left( 3 \right)\\2B + D = 1 \to \left( 4 \right)\end{array}\)

Find the values of A,B, C,D

Subtract Equation 1 from Equation 3

\(\begin{array}{l}2A + C - A - C = 2 - 1\\\therefore A = 1\end{array}\)

Put the value of\(A = 1\)in Equation 1

So we get,\(1 + C = 1 \Rightarrow C = 0\)

Subtract the Equation 2 from Equation 4

\(2B + D - B - D = 1 - 1 \Rightarrow B = 0\)

Put the value of\(B = 0\)in Equation 2

\(\therefore 0 + D = 1 \Rightarrow D = 1\)

Find the Integral

\(\begin{array}{c}\therefore \frac{{{x^3} + {x^2} + 2x + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}} = \frac{{Ax + B}}{{{x^2} + 1}} + \frac{{Cx + D}}{{{x^2} + 2}}\\\end{array}\)

Apply integration and substitute the values of A,B,C,D

\(\therefore \int {\frac{{{x^3} + {x^2} + 2x + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}} dx = \int {\left( {\frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 2}}} \right)} dx\)

Use,\(\int {\frac{{{F^'}\left( x \right)}}{{F\left( x \right)}}} dx = \ln |f\left( x \right)| + C\& \int {\frac{1}{{{x^2} + {a^2}}} = \frac{1}{a}} {\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c\)

\(\begin{array}{l}\therefore \int {\frac{{{x^3} + {x^2} + 2x + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}dx = \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}} } dx - \int {\frac{x}{{{x^2} + {{\left( {\sqrt 2 } \right)}^2}}}dx} \\\therefore \int {\frac{{{x^3} + {x^2} + 2x + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}dx = \frac{1}{2}\ln |{x^2} + 1| + \frac{1}{{\sqrt 2 }}} {\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + c\end{array}\)

Hence the value of the given integral is

\(\int {\frac{{{x^3} + {x^2} + 2x + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}dx = \frac{1}{2}\ln |{x^2} + 1| + \frac{1}{{\sqrt 2 }}} {\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + c\)

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