Originally Posted by

**madmartigano** I need to show that $\displaystyle \int {\frac{x}{{{x^2}+4}}dx}

=\frac{1}{2}\ln \left( {{x^2} + 4} \right) + c

$ using trig substitution.

I've tried the following:

$\displaystyle \begin{array}{l}

x = 2\tan \theta , dx = 2{\sec ^2}\theta d\theta \\

\\

\int {\frac{{4\tan \theta {{\sec }^2}\theta }}{{4{{\tan }^2}\theta + 4}}d\theta = \int {\frac{{\tan \theta {{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta = \int {\tan \theta d\theta = \ln \left| {\sec \theta } \right| + c} } } \\

\end{array}

$

$\displaystyle \theta = {\tan ^{ - 1}}\frac{x}{2}, \sec \theta = \frac{{\sqrt {{x^2} + 4} }}{2},

$

but $\displaystyle \int {\frac{x}{{{x^2} + 4}}dx \ne \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)} + c

$

Where am I going wrong?