# Thread: Integration Problem

1. ## Integration Problem

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 $\displaystyle -e^{8-6x} + C$

2. Hello,

It's all correct 3. 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:
$\displaystyle -\frac{1}{4}\sqrt{6x}+C$

4. 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 5. oops no you were right I did it wrong but thanks for pointing that out 6. the answer should be:

$\displaystyle \frac{3}{4}x-\frac{1}{2}\sqrt{6x}+C$

correct?

7. No, it looks like you forgot du=6 dx Otherwise, it's correct.

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