# Evaluate: \int_2^4\frac{\sqrt{\ln(9-x)}dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}

Evaluate:
${\int }_{2}^{4}\frac{\sqrt{\mathrm{ln}\left(9-x\right)}dx}{\sqrt{\mathrm{ln}\left(9-x\right)}+\sqrt{\mathrm{ln}\left(x+3\right)}}$
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zurilomk4
Let
$L={\int }_{2}^{4}\frac{\sqrt{\mathrm{ln}\left(9-x\right)}}{\sqrt{\mathrm{ln}\left(9-x\right)}+\sqrt{\mathrm{ln}\left(3+x\right)}}dx$
Now, use that
${\int }_{a}^{b}f\left(x\right)dx={\int }_{a}^{b}f\left(a+b-x\right)dx$
Then
$I={\int }_{2}^{4}\frac{\sqrt{\mathrm{ln}\left(3+x\right)}}{\sqrt{\mathrm{ln}\left(3+x\right)}+\sqrt{\mathrm{ln}\left(9-x\right)}}dx$
Add up these two integrals to get
$2I={\int }_{2}^{4}\frac{\sqrt{\mathrm{ln}\left(9-x\right)}+\sqrt{\mathrm{ln}\left(3+x\right)}}{\sqrt{\mathrm{ln}\left(9-x\right)}+\sqrt{\mathrm{ln}\left(3+x\right)}}dx$
Thus,
$I=1$
In order to prove (1), write the integral using another variable, say, t:
${\int }_{a}^{b}f\left(a+b-x\right)dx={\int }_{a}^{b}f\left(a+b-t\right)dt$
In the latter one, set $x=a+b-t$ and $dt=-dx$ and change the limits of integration to obtain
${\int }_{a}^{b}f\left(a+b-t\right)dt=-{\int }_{b}^{a}f\left(x\right)dx$
$={\int }_{a}^{b}f\left(x\right)dx$
Mason Hall