Question

Evaluate the Integral:\(\int\limits_0^1 {\frac{x}{{{x^2} + 4x + 13}}dx} \)

Short answer

The value of the given Integral can be given as:

\(\int\limits_0^1 {\frac{x}{{{x^2} + 4x + 13}}} dx = \frac{1}{2}\ln \frac{{18}}{{13}} - \frac{\pi }{6} + \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{3}{2}} \right)\)

Step-by-step solution

Rewrite the Integral

\(\begin{array}{l}\int\limits_0^1 {\frac{x}{{{x^2} + 4x + 13}}} dx = \int {\frac{x}{{{{\left( {x + 2} \right)}^2} + 9}} + \frac{{\left( {x + 2} \right) - 2}}{{{{\left( {x + 2} \right)}^2} + 9}}dx} \\\int\limits_0^1 {\frac{x}{{{x^2} + 4x + 13}}} dx = \int\limits_0^1 {\left( {\frac{{x + 2}}{{{{\left( {x + 2} \right)}^2} + 9}} - \frac{2}{{{{\left( {x + 2} \right)}^2} + 9}}} \right)} dx\\\end{array}\)

Substitution \(\alpha  = \left( {x + 2} \right)\) and \(d\alpha  = dx\) and find Integral

The limits of integration are:

If\(x = 0\)then\(\alpha = 2\)and if\(x = 1\)then\(\alpha = 3\)

So the above integral becomes,

\(\begin{array}{l}\int\limits_0^1 {\frac{x}{{{x^2} + 4x + 13}}} dx = \int\limits_0^1 {\left( {\frac{{x + 2}}{{{\alpha ^2} + 9}} - \frac{2}{{{\alpha ^2} + 9}}} \right)} dx\\ = \frac{1}{2}\int\limits_2^3 {\frac{{2x}}{{{\alpha ^2} + 9}}dx - 2\int\limits_2^3 {\frac{1}{{{\alpha ^2} + {3^2}}}} } dx\\\end{array}\)

Use, \(\begin{array}{l}\int {\frac{{{f^'}\left( x \right)}}{{f\left( x \right)}}} dx = \ln |\left( {f\left( x \right)} \right)| + c\& \int {\frac{{dx}}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c\\\therefore \int\limits_0^1 {\frac{x}{{{x^2} + 4x + 13}}} dx = \frac{1}{2}\left( {\ln |{\alpha ^2} + 9|} \right)_2^3 - 2\left( {\frac{1}{3}{{\tan }^{ - 1}}\frac{\alpha }{3}} \right)_2^3\end{array}\)

Apply the limits and obtain the final answer

Hence,

\(\begin{array}{l}\int\limits_0^1 {\frac{x}{{{x^2} + 4x + 13}}} dx = \frac{1}{2}\left( {\ln 18 - \ln 13} \right) - \frac{2}{3}\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\frac{3}{2}} \right)\\\end{array}\)

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