# How do you integrate int x/sqrt(x^2+9)dx?

Sep 17, 2015

substitute x = $3 \tan \theta$
$\left({x}^{2} + 9\right)$=$9 {\tan}^{2} \theta + 9$= $9 {\sec}^{2} \theta$
${\left({x}^{2} + 9\right)}^{\frac{1}{2}}$ = $3 \sec \theta$
$\mathrm{dx} = 3 {\sec}^{2} \theta d \theta$
substiuting
$\int$$\frac{3 \tan \theta}{3 \sec \theta}$ $3 {\sec}^{2} \theta d \theta$
=$3 \int$ $\sec \theta \tan \theta d \theta$
=$3 \sec \theta + c$
sectheta = sqrt $\left(1 + {\tan}^{2} \theta\right)$=(root1+(x/3)^2)
$\frac{\sqrt{9 + {x}^{2}}}{3}$
=>$3 \sec \theta$=sqrt(9+x^2
the final answer is $\sqrt{9 + {x}^{2}} + c$
it can be done in other way
$x i s \mathrm{de} r r i v a t i v e o f {x}^{2} / 2$
it can be wriiten as $d \frac{{x}^{2}}{2}$
so $\int \frac{x}{\sqrt{{x}^{2} + 9}} \mathrm{dx}$ bexmes $\frac{1}{2} \int \frac{1}{\sqrt{{x}^{2} + 9}} d \left({x}^{2}\right)$
=$\frac{1}{2} \left(2 \sqrt{{x}^{2} + 9}\right)$+$c$
=$\sqrt{{x}^{2} + 9} + c$

Sep 18, 2015

$\sqrt{{x}^{2} + 9} + C$

#### Explanation:

substitution:

${x}^{2} + 9 = t \implies 2 x \mathrm{dx} = \mathrm{dt} \implies x \mathrm{dx} = \frac{\mathrm{dt}}{2}$

$\int \frac{x}{\sqrt{{x}^{2} + 9}} \mathrm{dx} = \int \frac{\mathrm{dt}}{2} \frac{1}{\sqrt{t}} = \frac{1}{2} \int {t}^{- \frac{1}{2}} \mathrm{dt} =$

$= \frac{1}{2} {t}^{\frac{1}{2}} / \left(\frac{1}{2}\right) + C = \sqrt{t} + C = \sqrt{{x}^{2} + 9} + C$