# Evaluate the integrals \int_0^{\ln9}e^\theta(e^\theta-1)^{1/2}d\theta

Evaluate the integrals
${\int }_{0}^{\mathrm{ln}9}{e}^{\theta }{\left({e}^{\theta }-1\right)}^{\frac{1}{2}}d\theta$
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wheezym
We have to evaluate the integral
${\int }_{0}^{\mathrm{ln}9}{e}^{\theta }{\left({e}^{\theta }-1\right)}^{\frac{1}{2}}d\theta$
let ${e}^{\theta }-1==t$
on differentiating with respect to t we get
${e}^{\theta }d\theta =dt$
$d\theta =\frac{dt}{{e}^{\theta }}$
now the limits
$t={e}^{\theta }-1$
$t={e}^{\mathrm{ln}9}-1$
$t=9-1$
$t=8$
also
$t={e}^{0}-1$
$t=1-1$
$t=0$
now the definite integral
${\int }_{0}^{8}{e}^{\theta }{t}^{\frac{1}{2}}\frac{dt}{{e}^{\theta }}$
${\int }_{0}^{8}{t}^{\frac{1}{2}}dt$
$⇒{\left[\frac{{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_{0}^{8}$
$⇒\frac{2}{3}{\left[{t}^{\frac{3}{2}}\right]}_{0}^{8}$
$⇒\frac{2}{3}\left[{8}^{\frac{3}{2}-0}\right]$
$⇒\frac{2}{3}\left[{2}^{3×\frac{3}{2}}\right]$
$⇒\frac{{2}^{\frac{11}{2}}}{3}=\frac{32×{2}^{\frac{1}{2}}}{3}$
therefore
${\int }_{0}^{\mathrm{ln}9}{e}^{\theta }{\left({e}^{\theta }-1\right)}^{\frac{1}{2}}d\theta =\frac{32×{2}^{\frac{1}{2}}}{3}$