Answer
(1/2)ln|$x^{2}$+4|+C
Work Step by Step
-- Trigonometry substitution method:
The denominator has the form ($x^{2}$+$a^{2}$), so let's replace x by a·tan θ. In this way, ($x^{2}$+$a^{2}$) is converted to ($a^{2}$$tan^{2} θ$+$a^{2}$), which is factored into $a^{2}$(1+$tan^{2} θ$). Then we apply the identity 1+$tan^{2} θ$=$sec^{2} θ$. In this problem, a=2, so we replace x by 2tan θ, and dx=2$sec^{2} θ$dθ
$\int$$\frac{x}{x^{2}+4}$dx= $\int$$\frac{2tan θ}{(2tan θ)^{2}+4}$(2$sec^{2} θ$dθ)
= $\int$$\frac{2tan θ}{4(1+tan^{2} θ)}$(2$sec^{2} θ$dθ)
= $\int$$\frac{2tan θ}{4sec^{2} θ}$(2$sec^{2} θ$dθ)
= $\int$tan θdθ
=-ln|cosθ|+C
Here, we convert θ back to x by using the trigonometry relationship shown by the image below.
The expression= -ln |$\frac{2}{√ (x^{2}+4)}$|+C
=ln$ |\frac{2}{ √(x^{2}+4)}|^{-1}$+C
= ln|√($x^{2}$+4)|-ln2+C
=(1/2)ln|$x^{2}$+4|+C'
-- Substitution with u method:
As u=$x^{2}$+4, du=2xdx
So $\int$$\frac{x}{x^{2}+4}$dx= $\int$$\frac{1}{2u}$du=(1/2)ln|u|+C=(1/2)ln|$x^{2}$+4|+C
