Evaluate the ∫e^z (sq rt of 1+e^z) dz
Solution.
Make a replacement:
t=1+ez,t≥0,t=\sqrt{1+e^z}, t\geq0,t=1+ez,t≥0, from here t2=1+ez,then 2tdt=ezdz.t^2=1+e^z, \text{then } 2tdt=e^zdz.t2=1+ez,then 2tdt=ezdz.
We will have
∫t⋅2tdt=∫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.∫t⋅2tdt=∫2t2dt==23t3+C=32(1+ez)23+C.
Answer.
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