Question

Integral of (x+2)*sqrt(x-4) dx

I know the answer is 2/5*(x-4)^3/2 * (x+6) but I don't know how to get to it. U=4, du/dx= 1, but what's du=?

Answer 1

ask a human caculater

Answer 2

Uhhh that's not really helpful

Answer 3

u = x-4 ----> x = u+4 or x+2 = u+6
du = dx

\[\rightarrow \int\limits (u+6) \sqrt{u} du\]

Answer 4

split it into 2 integrals after distributing...
\[\int\limits u \sqrt{u} + \int\limits 6 \sqrt{u}\]

Answer 5

put \[\sqrt{x-4}=t,x-4=t^2,x=t^2+4,dx=2tdt\]
\[I=\int\limits \left( t^2+4+6 \right)t*2tdt=2\int\limits \left( t^4+10t^2 \right)dt=?\]

Answer 6

correction write 2 in place of 6 and 6 in place of 10