Evaluate the Integral: $\int\ e^{x}(1+e^x)^{\frac{1}{2}}\ dx$ Evaluate the Integral: $\int\ e^{x}(1+e^x)^{\frac{1}{2}}\ dx$
$u=1+e^x$
$du=e^x\ dx$
$\int\ u^{\frac{1}{2}}\ du$
$u+c=1+e^x+c$
 A: Your change of variable is OK, $ u=e^x$, $du=e^xdx$, giving
$$\int e^{x}(1+e^x)^{\frac12}\ dx=\int (1+u)^{\frac12}\ du=\frac23(1+u)^{3/2}+C=\frac23(1+e^x)^{3/2}+C.$$
A: $$\int\ e^{3}(1+e^x)^{\frac{1}{2}}\ dx=$$
$$e^3\int\sqrt{e^x+1} dx=$$
(substitute $u=e^x$ and $du=e^xdx$):
$$e^3\int\frac{\sqrt{u+1}}{u} du=$$
(substitute $s=\sqrt{u+1}$ and $ds=\frac{1}{2\sqrt{u+1}}du$):
$$2e^3\int\frac{s^2}{s^2-1} ds=$$
$$2e^3\int\left(-\frac{1}{2(s+1)}+\frac{1}{2(s-1)}+1\right)ds=$$
$$-e^3\int \frac{1}{s+1}ds+e^3\int \frac{1}{s-1}ds+2e^3\int1ds=$$
(substitute $p=s+1$ and $dp=ds$):
$$-e^3\int \frac{1}{p}dp+e^3\int \frac{1}{s-1}ds+2e^3\int1ds=$$
$$-e^3\ln(p)+e^3\int \frac{1}{s-1}ds+2e^3\int1ds=$$
(substitute $q=s-1$ and $dw=ds$):
$$-e^3\ln(p)+e^3\int \frac{1}{w}dw+2e^3\int1ds=$$
$$-e^3\ln(p)+e^3\ln(w)+2e^3\int1ds=$$
$$-e^3\ln(p)+e^3\ln(w)+2e^3s+C=$$
$$-e^3\ln(p)+e^3\ln(s-1)+2e^3s+C=$$
$$-e^3\ln(\sqrt{u+1}+1)+e^3\ln(\sqrt{u+1}-1)+2e^3\sqrt{u+1}+C=$$
$$-e^3\ln(\sqrt{e^x+1}+1)+e^3\ln(\sqrt{e^x+1}-1)+2e^3\sqrt{e^x+1}+C=$$
$$2e^3\left(\sqrt{e^x+1}-\tanh^{-1}\left(\sqrt{e^x+1}\right)\right)+C$$
A: $$\int e^{x}(1+e^x)^{\frac{1}{2}}\ dx$$
Using $u$-substitution, we have
$$ u=e^x+1$$
$$ du=e^x\ dx$$
So now
$$\int u^{\frac{1}{2}}\ du=\frac{u^{\frac12+1}}{\frac12+1}+C$$
$$=\frac{u^{\frac12+\frac22}}{\frac12+\frac22}+C=\frac{u^{\frac32}}{\frac32}+C$$
$$=\frac23 u^{\frac32}+C=\frac23 \left(e^x+1\right)^{\frac32}+C$$
