Question

CALCULUS : Find the value of the Integral:

Answer 1

\[\int\limits_{}^{}x \sqrt{x^2-9}\]

Answer 2

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

Answer 3

@ayoubEpst Please use the EQUATION EDITOR

Answer 4

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

Answer 5

@ayoubEpst I have wolfram alpha too

Answer 6

no no

Answer 7

I don't need the answer

Answer 8

yes i can explain

Answer 9

I hope you can

Answer 10

\[\int\limits_{?}^{?} u'u^n\]

Answer 11

you know the integral of that kind ??

Answer 12

no i don't

Answer 13

I would just make a u-substitution here

make u = x^2 - 9
so du = 2x dx

and du/2 = xdx

Now you have

\[\large \frac{1}{2}\int \sqrt{u} du\]

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

that's why I'm sutcked

Answer 18

\[u'=x ; u=(x^2-9)^1/2\]

Answer 19

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

Answer 20

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