Question

: Find the integral of

\(I = \int {\frac{{\left( {x - 3} \right)dx}}{{{{\left( {{x^2} + 2x + 4} \right)}^2}}}} \)

Short answer

Multiply and divide by 2 and thus split into parts thus simplifications become easier

Step-by-step solution

Step 1: Substitution

\(I = \frac{1}{2}\int {\frac{{2\left( {x - 3} \right)dx}}{{{{\left( {{x^2} + 2x + 4} \right)}^2}}}} \)

=\(\frac{1}{2}\int {\frac{{2x - 6dx}}{{{{\left( {{x^2} + 2x + 4} \right)}^2}}}} \)

\( = \frac{1}{2}\left( {\int {\frac{{2x + 2}}{{\left( {{x^2} + 2x + 4} \right)}}dx - \int {\frac{{8dx}}{{{{\left( {{x^2} + 2x + 4} \right)}^2}}}} } } \right)\)

\( = \frac{1}{2}\left( {{I_1} - {I_2}} \right)\)

Simplification the \({I_1}\) part

\({I_1} = \int {\frac{{2x + 2}}{{{{\left( {{x^2} + 2x + 4} \right)}^2}}}dx} \)

Put \({x^2} + 2x + 4 = t\)

\(2x + 2dx = dt\)

\(\therefore {I_1} = \int {\frac{{dt}}{{{t^2}}} = \int {{t^{ - 2}}dt} } \)

\( = \frac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + c\)

\( = \frac{{{t^{ - 1}}}}{{ - 1}} + c = \frac{{ - 1}}{t} + c\)

\(\therefore {I_1} = \frac{{ - 1}}{{{x^2} + 2x + 4}} + c\)

Simplification the \({I_2}\) part

\({I_2} = \int {\frac{8}{{{{\left( {{x^2} + 2x + 4} \right)}^2}}}} dx\)

\( = 8\int {\frac{1}{{{{\left( {{x^2} + 2x + 1 + 3} \right)}^2}}}} dx\)

\( = 8\int {\frac{1}{{{{\left( {{{\left( {x + 1} \right)}^2} + 3} \right)}^2}}}} dx\)

Put \(x + 1 = u\)

\(dx = du\)

\( = 8\int {\frac{{du}}{{{{\left( {{u^2} + 3} \right)}^2}}}} \)

Apply reduction formula

\(\int {\frac{1}{{{{\left( {a{u^2} + b} \right)}^n}}}du = \frac{{2n - 3}}{{2b\left( {n - 1} \right)}}\int {\frac{{du}}{{{{\left( {a{u^2} + b} \right)}^{n - 1}}}} + \frac{u}{{2b\left( {n + 1} \right){{\left( {a\left( {{u^2} + b} \right)} \right)}^{n - 1}}}}} } \)

\(\therefore a = 1,b = 3,n = 2\)

\( = 8\left( {\frac{u}{{6\left( {{u^2} + 3} \right)}} + \frac{1}{6}\int {\frac{1}{{{u^2} + 3}}du} } \right)\)

\( = 8\left( {\frac{u}{{6\left( {{u^2} + 3} \right)}} + \frac{1}{6}.\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right)} \right) + c\)

\( = 8\left( {\frac{u}{{6\left( {{{\left( {x + 1} \right)}^2} + 3} \right)}} + \frac{1}{{6\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 3 }}} \right)} \right) + c\)

\( = \frac{8}{6}\left( {\frac{{x + 1}}{{{x^2} + 2x + 4}} + \frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 3 }}} \right)} \right) + c\)

Substitution of \({I_1}\) and \({I_2}\)

\(\therefore I = \frac{1}{2}\left( {{I_1} - {I_2}} \right)\)

\( = \frac{1}{2}\left( {\frac{{ - 1}}{{{x^2} + 2x + 4}} + \frac{4}{3}\left( {\frac{{x + 1}}{{{x^2} + 2x + 4}} + \frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 3 }}} \right)} \right)} \right) + c\)

\( = \frac{{ - 1}}{{2\left( {{x^2} + 2x + 4} \right)}} + \frac{2}{3}.\frac{{\left( {x + 1} \right)}}{{\left( {{x^2} + 2x + 4} \right)}} + \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 3 }}} \right) + c\)

\(\therefore \)Final answer:\(I = \int {\frac{{x - 3}}{{{{\left( {{x^2} + 2x + 4} \right)}^2}}}dx} \)

\( = \frac{{ - 1}}{{2\left( {{x^2} + 2x + 4} \right)}} + \frac{2}{3}\frac{{\left( {x + } \right)1}}{{\left( {{x^2} + 2x + 4} \right)}} + \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 3 }}} \right) + c\)