Question

Evaluate the integral: \(\int {\frac{{(x - 3)}}{{{{({x^2} + 2x + 4)}^2}}}} \)

Short answer

Multiply and divide by 2 and thus split into parts thus simplification becomes easier

Step-by-step solution

Substitution

\(\begin{array}{l}I = \frac{1}{2}\int {\frac{{2(x - 3)}}{{{{({x^2} + 2x + 4)}^2}}}} dx\\I = \frac{1}{2}\int {\frac{{(2x - 6)}}{{{{({x^2} + 2x + 4)}^2}}}dx} \\I = \frac{1}{2}\int {\frac{{(2x + 2)}}{{{{({x^2} + 2x + 4)}^2}}}} dx - \frac{1}{2}\int {\frac{8}{{{{({x^2} + 2x + 4)}^2}}}dx} \\I = \frac{1}{2}({I_1} - {I_1})\end{array}\)

  Simplification\({I_1}\)

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

Put \(\begin{array}{l}{x^2} + 2x + 4 = t\\2x + 2 = dt\end{array}\)

\(\begin{array}{l}{I_1} = \int {\frac{{dt}}{{{t^2}}}} = \int {{t^{ - 2}}dt} \\{I_1} = \frac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + c\\{I_1} = \frac{{{t^{ - 1}}}}{{ - 1}} + c\\{I_1} = \frac{{ - 1}}{t} + c\\{I_1} = \frac{{ - 1}}{{{x^2} + 2x + 4}} + c\end{array}\)

  Simplification\({I_2}\)

\(\begin{array}{l}{I_2} = 8\int {\frac{1}{{{{({x^2} + 2x + 4)}^2}}}dx} \\{I_2} = 8\int {{{({x^2} + 2x + 4)}^{ - 2}}dx} \\{I_2} = 8\frac{{{{({x^2} + 2x + 4)}^{ - 2 + 1}}}}{{ - 2 + 1}}(2x + 2) + c\\{I_2} = 8\frac{{{{({x^2} + 2x + 4)}^{ - 1}}}}{{ - 1}}(2x + 2) + c\\{I_2} = 8\frac{{ - 1}}{{({x^2} + 2x + 4)}}(2x + 2) + c\end{array}\)

  Find the integral \(I\)

\(\begin{array}{l}I = \frac{1}{2}({I_1} - {I_1})\\I = \frac{1}{2}\left( {\frac{{ - 1}}{{{x^2} + 2x + 4}} + 8\frac{1}{{({x^2} + 2x + 4)}}(2x + 2)} \right) + c\\I = \left( {\frac{{ - 1}}{{2{x^2} + 4x + 8}} + \frac{4}{{({x^2} + 2x + 4)}}(2x + 2)} \right) + c\end{array}\)

Hence, the solution of the given integral will be:\(I = \left( {\frac{{ - 1}}{{2{x^2} + 4x + 8}} + \frac{4}{{({x^2} + 2x + 4)}}(2x + 2)} \right) + c\).