(integral symbol) (7e^(2x)-3e^-x) dx
Thanks guys i dont know how to do it with the e's
Break it up first
$\displaystyle 7\int{e^{2x}}-3\int{e}^{-x}$
In the first let $\displaystyle u_1=2x$, then $\displaystyle \frac{1}{2}du_1=dx$
In the second let $\displaystyle u_2=-x$, then $\displaystyle -du_2=dx$
(Iwrite subscripts to avoid confusion, I should have used u AND v, but too late now)...
and we have
$\displaystyle \frac{7}{2}\int{e^{u_1}}du_1+3\int{e^{u_2}}du_2$
The integral of $\displaystyle e^u:e^u$
Therefore (subbing back for u)
$\displaystyle =\frac{7}{2}e^{2x}+3e^{-x}+C$