Question #162050

Evaluate the ∫e^z (sq rt of 1+e^z) dz


Expert's answer

Solution.

ez1+ezdz.\int e^z\sqrt{1+e^z}dz.

Make a replacement:

t=1+ez,t0,t=\sqrt{1+e^z}, t\geq0, from here t2=1+ez,then 2tdt=ezdz.t^2=1+e^z, \text{then } 2tdt=e^zdz.

We will have

t2tdt=2t2dt==2t33+C=23(1+ez)32+C.\int t\cdot 2tdt=\int 2t^2dt=\newline= 2\frac{t^3}{3}+C= \frac{2}{3}(1+e^z)^{\frac{3}{2}}+C.

Answer.

23(1+ez)32+C.\frac{2}{3}(1+e^z)^{\frac{3}{2}}+C.
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