Integrate:
$\displaystyle
6(e^{4-3x})^2
$
Is it correct if i do this: $\displaystyle 6(e^{8-6x})$ and then solve?
and then let u = 8-6x and du=-6
then the answer comes to $\displaystyle -e^{8-6x} + C$
thank you, here is another one (I have a test on Friday so I'm trying to do as many problems as I can)
Integrate: $\displaystyle \frac{3}{4}-\frac{1}{\sqrt{6x}}$
I let u = 6x and du = 6dx
the answer I end up with is:
$\displaystyle
-\frac{1}{4}\sqrt{6x}+C
$
Hello,
And what about integrating the constant ? The integral of 1 is x, the integral of 3/4 will be...
But according to your answer, would the integral rather be $\displaystyle \int \frac 34\times \frac{1}{\sqrt{6x}} ~dx$ ? In which case, it's correct - again