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Evaluate the integral. \displaystyle\int \frac { 4 x + 4 } { x ^ { 4 } + x ^ { 3 } + 2 x ^ { 2 }...

Question:

Evaluate the integral.

{eq}\displaystyle\int \frac { 4 x + 4 } { x ^ { 4 } + x ^ { 3 } + 2 x ^ { 2 } } d x {/eq}

Integral using Substitution:

Let us consider the integral {eq}I = \int {f(g(x))} g'(x)dx {/eq}, where {eq}f(x) {/eq} and {eq}g(x) {/eq} are real valued functions of {eq}x {/eq}.

Let us assume the substitution {eq}z = g(x) \Rightarrow dz = g'(x)dx {/eq}

After substitution:

{eq}I = \int {f(z)} dz {/eq}, which is easily solvable.

Answer and Explanation:

{eq}\eqalign{ I& = \int {\frac{{4x + 4}}{{{x^4} + {x^3} + 2{x^2}}}} dx \cr & = \int {\frac{{4x + 4}}{{{x^2}\left( {{x^2} + x + 2} \right)}}} dx \cr & = \int {\left( {\frac{{ - x - 3}}{{{x^2} + x + 2}} + \frac{2}{{{x^2}}} + \frac{1}{x}} \right)} dx\,\,\,\,\,\,\,\left[ {{\text{Using the partial fraction: }}\frac{{4x + 4}}{{{x^4} + {x^3} + 2{x^2}}} = \frac{{Ax + B}}{{{x^2} + x + 2}} + \frac{C}{{{x^2}}} + \frac{D}{x}} \right] \cr & = \int {\left( {\frac{{ - x - 3}}{{{x^2} + x + 2}} + \frac{2}{{{x^2}}} + \frac{1}{x}} \right)} dx \cr & = \int {\left( {\frac{{ - x - 3}}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = \int {\left( {\frac{{ - x}}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{{ - 3}}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = - \frac{1}{2}\int {\left( {\frac{{2x + 1 - 1}}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{{ - 3}}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = - \frac{1}{2}\int {\left( {\frac{{2x + 1}}{{{x^2} + x + 2}}} \right)} dx + \frac{1}{2}\int {\left( {\frac{1}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{{ - 3}}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = - \frac{1}{2}\int {\left( {\frac{{2x + 1}}{{{x^2} + x + 2}}} \right)} dx - \frac{5}{2}\int {\left( {\frac{1}{{{x^2} + x + 2}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = - \frac{1}{2}\int {\left( {\frac{{2x + 1}}{{{x^2} + x + 2}}} \right)} dx - \frac{5}{2}\int {\left( {\frac{1}{{{x^2} + 2\left( {\frac{1}{2}} \right)x + \frac{1}{4} - \frac{1}{4} + 2}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = - \frac{1}{2}\int {\left( {\frac{{2x + 1}}{{{x^2} + x + 2}}} \right)} dx - \frac{5}{2}\int {\left( {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{7}{4}}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = - \frac{1}{2}\int {\left( {\frac{{d\left( {{x^2} + x + 2} \right)}}{{{x^2} + x + 2}}} \right)} - \frac{5}{2}\int {\left( {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\sqrt {\frac{7}{4}} } \right)}^2}}}} \right)} dx + \int {\left( {\frac{2}{{{x^2}}}} \right)} dx + \int {\left( {\frac{1}{x}} \right)} dx \cr & = - \frac{1}{2}\ln \left( {{x^2} + x + 2} \right) - \frac{5}{2}\frac{1}{{\left( {\sqrt {\frac{7}{4}} } \right)}}{\tan ^{ - 1}}\frac{{\left( {x + \frac{1}{2}} \right)}}{{\left( {\sqrt {\frac{7}{4}} } \right)}} - \frac{2}{x} + \ln \left( x \right) + c\backslash \cr & = - \frac{1}{2}\ln \left( {{x^2} + x + 2} \right) - \frac{5}{{\sqrt 7 }}{\tan ^{ - 1}}\frac{{2x + 1}}{{\sqrt 7 }} - \frac{2}{x} + \ln \left( x \right) + c \cr} {/eq}

Where {eq}c {/eq} is the integrating constant.


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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