# I Integration technique

1. Jul 13, 2016

### albalaka

I have been given the problem ∫√(ex-3)

and we must use the substitution u = √(ex-3)

I can start it off with u = √(ex-3)
and du = exdx/2u

and what Ive been trying is to complete the square and go towards 2 ∫ u2du/((u2+4) -1)

But im not getting towards the answer, either Im doing something wrong in the middle, or my approach is wrong.

the answer given is 2√(ex-3) - 2 √3 arctan(ex-3)/√3) +c

any help would really be appreciated, ive searched online trying to find a pattern, but i just cant figure it out

2. Jul 13, 2016

### malawi_glenn

u = sqrt(e^x - 3)

du/dx = e^x/(2*u)

now you can solve for e^x in terms of u: e^x = u^2 + 3

can you take it from here?

3. Jul 13, 2016

### albalaka

Thats where I can get up to, I have 2 ∫ u2du/(u2 + 3)

I just cant figure out where to go from there

4. Jul 13, 2016

### malawi_glenn

integrate by parts

or write the nominator in the integrand 2u^2 as 2(u^2+3-3)

5. Jul 13, 2016

### pasmith

I would suggest $u = \sqrt{3}\tan\theta$.

6. Jul 13, 2016

### malawi_glenn

totally overkill :) just add and subtract 3 from the nominator

7. Jul 13, 2016

### pasmith

But how do you know what $-2\int \frac 3{3 + u^2}\,du$ is?

Last edited: Jul 13, 2016
8. Jul 13, 2016

### Math_QED

Well, a trig substitution is not nessecary there ;)

9. Jul 13, 2016

### malawi_glenn

I thought the integral of 1/(x^2+1) was standard to know