Integration?

Question

5/[(x-1)*sqrt(x^2-2x-24)]

Answer 1

Ok, this is not an easy integration, here you have to apply substituions, first :

integrate(5/[(x-1)*sqrt(x^2-2x-24)]dx

Let's make : x-1 = t >>> dt = dx

x^2 - 2x - 24 = (x-6)(x+4) = (t-5)(t+5) = t^2 - 25

integrate(5dt / [t*sqrt(t^2 - 25)]

5*integrate(t^-1*(t^2 - 25)^-1/2)dt

Now, you have a better integration, now apply another substitution :

t^2 - 25 = u^2

2tdt = 2udu >>> dt = udu / t

t^2 = u^2 + 25

Let's replace the values, step by step :

5*integrate(u^-1*t^-1dt)

5*integrate(u^-1*u*du / t^2)

5*integrate( du / u^2 + 25)

Now, there is something we can integrate :

integrate( du / u^2 + 25) = arctg(u/5)

u = sqrt(t^2 - 25)

and t = x-1

u = sqrt((x-1)^2 - 25)

The final result will be :

arctg(sqrt((x-1)^2 - 25 / 5 )

THAT'S IT

Hope that might help you

Answer 2

Integration via areas. The functionality is a made from 2 applications: (x) * (cosx) call one in all them "u" and the different "dv" decide on the "u" so as that its spinoff simplifies u = x du = a million*dx dv = cosx v = sinx this is for the formulation setup, it somewhat is u*v - necessary (vdu) =xsinx - necessary(sinx * a million * dx) =xsinx - (-cosx) + C = xsinx + cosx + C

Answer 3

...??? Integrate it on what bounds?