Question

Integration:

Answer 1

\[\Large \int x^5 \sqrt{x^2+4} dx\]

Answer 2

This looks trigonometric as 4 can be written as 2^2

Answer 3

I haven't had much practice with trig sub so I could never use that sub too well.

Answer 4

x^2 x^2 sqrt(x^2+4) x dx

u = x^2 + 4

Answer 5

I'd be surprised if that actually got the answer we needed..

but \(\Large \frac{1}{2} \int (u+4)^2~\sqrt{u}~du\)?

Answer 6

it's u - 4

Answer 7

Whoops, the + and - are really close together on this keyboard :P

But now we just distribute and simplify?

Answer 8

yes

Answer 9

\[I=\int\limits x^5\sqrt{x^2+4}dx\]
\[=\int\limits \left(x^2 \right)^2\sqrt{x^2+4}~x~dx\]
\[Put~\sqrt{x^2+4}=t,x^2=t^2-4,2x~dx=2~t~dt,x~dx=t~dt\]
\[I=\int\limits \left( t^2-4 \right)^2~t~t~dt=\int\limits \left( t^4-8t^2+16 \right)t^2~dt=\int\limits \left( t^6-8 t^4+16~t^2 \right)dt\]
\[=\frac{ t^7 }{ 7 }-\frac{ 8~t^5 }{ 5 }+\frac{ 16~t^3 }{ 3 }+c\]
replace the value of t