How do you integrate $\int \frac{x}{\sqrt{x^{2} + 9}} d x$ $int x/sqrt(x^2+9)dx$ ?

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How do you integrate $\int \frac{x}{\sqrt{x^{2} + 9}} d x$ $int x/sqrt(x^2+9)dx$ ?

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Answer 1 · 2015-09-17T08:20:03

substitute x = $3 tan θ$ $3 tantheta$
$(x^{2} + 9)$ $(x^2+9)$ = $9 {tan}^{2} θ + 9$ $9tan^2theta+9$ = $9 {sec}^{2} θ$ $9sec^2theta$
${(x^{2} + 9)}^{\frac{1}{2}}$ $(x^2+9)^(1/2)$ = $3 sec θ$ $3sec theta$
$d x = 3 {sec}^{2} θ d θ$ $dx = 3 sec^2theta d theta$
substiuting
$\int$ $int$ $\frac{3 tan θ}{3 sec θ}$ $(3tantheta)/(3sec theta)$ $3 {sec}^{2} θ d θ$ $3 sec^2theta d theta$
= $3 \int$ $3int$ $sec θ tan θ d θ$ $sectheta tanthetad theta$
= $3 sec θ + c$ $3sectheta+c$
$sec θ = \sqrt{}$ $sectheta = sqrt$ $(1 + {tan}^{2} θ)$ $(1+tan^2theta)$ = $(\sqrt[]{}$ $(root$ $1 + {(\frac{x}{3})}^{2})$ $1+(x/3)^2)$
$\frac{\sqrt{9 + x^{2}}}{3}$ $sqrt(9+x^2)/3$
=> $3 sec θ$ $3sectheta$ = $\sqrt{9 + x^{2}}$ $sqrt(9+x^2$
the final answer is $\sqrt{9 + x^{2}} + c$ $sqrt(9+x^2)+c$
it can be done in other way
$x i s d e r r i v a t i v e o f \frac{x^{2}}{2}$ $x is derrivative of x^2/2$
it can be wriiten as $d \frac{x^{2}}{2}$ $d(x^2)/2$
so $\int \frac{x}{\sqrt{x^{2} + 9}} d x$ $int x/sqrt(x^2+9)dx$ bexmes $\frac{1}{2} \int \frac{1}{\sqrt{x^{2} + 9}} d (x^{2})$ $1/2int1/sqrt(x^2+9)d(x^2)$
= $\frac{1}{2} (2 \sqrt{x^{2} + 9})$ $1/2(2sqrt(x^2+9))$ + $c$ $c$
= $\sqrt{x^{2} + 9} + c$ $sqrt(x^2+9)+c$

Answer 2 · 2015-09-18T01:03:44

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

Explanation:

substitution:

$x^{2} + 9 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2}$ $x^2+9=t => 2xdx=dt => xdx=dt/2$

$\int \frac{x}{\sqrt{x^{2} + 9}} d x = \int \frac{d t}{2} \frac{1}{\sqrt{t}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t =$ $intx/sqrt(x^2+9)dx=intdt/2 1/sqrtt=1/2intt^(-1/2)dt=$

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

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