Having trouble with this integral ∫ sqrt(x^2 + 2x - 3) dx

Thanks for the help.

I have an attempt. I'm not sure if hyperbolic substitution is necessary but it's the only way I could think of that would work.

∫ sqrt(x^2 + 2x - 3) dx
= ∫ sqrt((x-1)^2 - 4) dx

let x - 1 = 2cosh(u); dx/du = 1/(2sinh(u)) => dx = 1/(2sinh(u)) du

=> ∫ sqrt(4cosh^2(u) - 4) dx
= ∫ sqrt(4sinh^2(u)) dx
= 2 ∫ sinh(u) dx
= 2 ∫ sinh(u) 1/(2sinh(u)) du
= ∫ du
= u + c

And I'm supposed to integrate between 1 and 2.

Also, is there a better way to solve this?

Thanks.

***PLEASE DON'T USE A TABLE.

Your sub is correct. But dx/du=2sinhu and the integral is
2 ∫2 sinh(u) sinh(u)du=4 ∫ sinh^2(u)du and you need to use
cosh(2u)=1+2sinh^2(u) to integrate this.
You have to change the limits using x+1=2coshu to give u=cosh^-1(1)
=0 and u=cosh^-1(3/2)=ln(3/2+sqrt(5/4)).
I don't know of any better way.

I use trigonometric substitutions for these problems

∫ sqrt((x+1)^2 - 4) dx
let x+1 = 2 secθ
dx = 2secθtanθ dθ

and substitue these values in your integral and you would get:

4∫tan^3θsecθ dθ
since the power of tan is odd, you make the substitution u=secθ and du=secθtanθ
you'll have the integral:
4∫ (u^2 -1) du
= 4/3 sec^3θ - 4secθ +C

you have to express it in terms of x, so the final answer would be:
4/3 (x+1/2)^3 - 4 (x+1/2)^3 + C

hope I helped :)

∫ sqrt(x^2 + 2x - 3) dx
= ∫ sqrt((x-1)^2 - 4) dx

wrong it is ∫ sqrt((x+1)^2 - 4) dx

now do it as u are doing


or u can remember the result for this integral
∫ root(x^2 - a^2) = x*root(x^2 - a^2)/2 - log|x + root(x^2 - a^2)| + c

so just substitute x+1 as t and apply the formula

Instead you can use this
∫sqrt(x^2 + 2x - 3) dx
= ∫ sqrt((x-1)^2 - 4) dx
= (x-1)sqrt(x^2 + 2x - 3)/2-(4/2)log(x^2 + 2x - 3)+c
then put x=2 and x=1 in the answer