Question

Evaluate the Integral:\(\int {\frac{{x + 4}}{{{x^2} + 2x + 5}}} dx\)

Short answer

The solution of the given Integral can be given as:

\(\int {\frac{{x + 4}}{{{x^2} + 2x + 5}}} dx = \frac{1}{2}\ln |{x^2} + 2x + 5| - \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) + c\)

Step-by-step solution

Rewrite the Integral

\(\begin{array}{l}\int {\frac{{x + 4}}{{{x^2} + 2x + 5}}dx = \int {\left( {\frac{{x + 1 + 3}}{{{{\left( {x + 1} \right)}^2} + 4}}} \right)} } = \int {\frac{{x + 1}}{{{{\left( {x + 1} \right)}^2} + 4}} + \frac{3}{{{{\left( {x + 1} \right)}^2} + 4}}} \\\end{array}\)

Use the substitution \(\alpha  = \left( {x + 1} \right)\) and find Integral

\(\begin{array}{l}\int {\frac{{x + 4}}{{{x^2} + 2x + 5}}} dx = \int {\left( {\frac{4}{{{\alpha ^2} + 4}} + \frac{3}{{{\alpha ^2} + 4}}} \right)} dx\\\int {\frac{{x + 4}}{{{x^2} + 2x + 5}}dx = \frac{1}{2}} \int {\frac{{2x}}{{{x^2} + {2^2}}}dx + 3\int {\frac{1}{{{x^2} + {2^2}}}} } dx\end{array}\)

Use,\(\int {\frac{1}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right)with,a = 2,\int {\frac{{df}}{f}} = \ln \left( f \right)with,f = {\alpha ^2} + 4\)

Then the integralis simplified to,

\(\begin{array}{l}\int {\frac{{x + 4}}{{{x^2} + 2x + 5}}dx = \frac{1}{2}\ln |{x^2} + 4| + 3\frac{1}{2}} {\tan ^{ - 1}}\frac{x}{2} + c\\ = \frac{1}{2}\ln |{x^2} + 4| + \frac{3}{2}{\tan ^{ - 1}}\frac{{x + 1}}{2} + c\\\therefore \int {\frac{{x + 4}}{{{x^2} + 2x + 5}}dx = \frac{1}{2}\ln |{x^2} + 2x + 5| + \frac{3}{2}} {\tan ^{ - 1}}\frac{{x + 1}}{2} + c\end{array}\)

Hence the value of the given integral is

\(\int {\frac{{x + 4}}{{{x^2} + 2x + 5}}dx = \frac{1}{2}\ln |{x^2} + 2x + 5| + \frac{3}{2}} {\tan ^{ - 1}}\frac{{x + 1}}{2} + c\)

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