Integrating √(ex-3) with Substitution: Step-by-Step Guide

[albalaka]

I have been given the problem ∫√(e^x-3)

and we must use the substitution u = √(e^x-3)

I can start it off with u = √(e^x-3)
and du = e^xdx/2u

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

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

the answer given is 2√(e^x-3) - 2 √3 arctan(e^x-3)/√3) +c

any help would really be appreciated, I've searched online trying to find a pattern, but i just can't figure it out

 

[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?

 

[albalaka]

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

I just can't figure out where to go from there

 

[malawi_glenn]

integrate by parts

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

 

[pasmith]

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

I just can't figure out where to go from there

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

 

[malawi_glenn]

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

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

 

[pasmith]

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

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

 

[member 587159]

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

Well, a trig substitution is not nessecary there ;)

 

[malawi_glenn]

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

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