CALCULUS : Find the value of the Integral:

Question

anonymous · Answer

$$\int\limits_{}^{}x \sqrt{x^2-9}$$

anonymous · Answer

int  x(x^2 -9)^ 1/2
1/3 (x^2-9)^3/2

anonymous · Answer

@ayoubEpst Please use the EQUATION EDITOR

anonymous · Answer

$$= 1/3 (x^2-9)\sqrt((x^2-9)) $$ +c

anonymous · Answer

@ayoubEpst I have wolfram alpha too

anonymous · Answer

no no

anonymous · Answer

I don't need the answer

anonymous · Answer

yes i can explain

anonymous · Answer

I hope you can

anonymous · Answer

$$\int\limits_{?}^{?} u'u^n$$

anonymous · Answer

you know the integral of that kind ??

anonymous · Answer

no i don't

johnweldon1993 · Answer

I would just make a u-substitution here

make u = x^2 - 9
so   du = 2x dx

and    du/2 = xdx

Now you have

$$\large \frac{1}{2}\int \sqrt{u} du$$

anonymous · Answer

@ayoubEpst You're method is good but I just want to know all the fomula's from the table

anonymous · Answer

so you want that i give you the table ???

anonymous · Answer

Sorry for my post but the real question for this problem was to be solved by using the integration by parts method so....

anonymous · Answer

that's why I'm sutcked

anonymous · Answer

$$u'=x ; u=(x^2-9)^1/2$$

anonymous · Answer

@ayoubEpst I know all about the integration by parts method but I'm stucked by a point further

anonymous · Answer

@ayoubEpst continue the integration and you'll see what I'm talking about