calc problem?

Question

int(r^3/sqrt(4+r^2))dr on the interval [0,1] using integration by parts

I've tried setting r^3=u =>du=3r^2 and setting dv=1/sqrt(4+r^2)
=>v=ln(r+sqrt(4+r^2))
This leads to the following:
int(r^3/sqrt(4+r^2))=(r^3)(ln(r+sqrt(4+r^2))) - int((3r^2)(ln(r+sqrt(4+r^2))))dr
This is the point I'm stuck at. Do I integrate by parts again? If so, what are my u and dv?

Answer 1

hi vooooooo
It is you answer;
∫ r^3 / (√(4+r^2)) dr = (1/3) ( x^2 - 8) (√x^2 +4) +c

Good Luck

Answer 2

the whole point of integrating by parts is to find a shortcut, in this case sqrt(4+r^2), is missing an r to be easily integrated, so try this:
u= r^2 , dv=u/sqrt(4+r^2)dr... hence du=2rdr v=(4+r^2)^(1/2)
so your problem reduces to:

uv - int(2r*sqrt(4+r^2)dr) = uv-2/3*(4+r^2)^(3/2) voila!

Answer 3

I do not thing you need to go by parts

realise that r^3 = r^2.r
now +tan^2 t = sec^2 t

put r= 2 tan t
dr/dt = 2 sec^t

r^3/sqrt(4+r^2) dr
= ((8 tan^3 t)/sqrt(4+4tan^2 t)) *(2 sec^t) dt
= (8 tan^3 t)/(2 sec t) *2 sec^t dt
= 8 tan^3 t sec t dt
= 8 sin^3 t /cos^4 t dt
= 8(1-cos ^2 t) sint/cos^4 t dt
now put cos t = y
sint dt = dy
we get
= -8(1-y^2)/y^4 dy
this is not integrable
you can now put the value and find the value

Answer 4

yo yo i got it, i think:

so u=r^3 and du=3r^2
dv=(-sqrt(4+r^2)) and v=2[+sqrt(4+r^2)]

so uv-(int)v du = (2r^3) (sqrt(4+r^2) -
(int) 2*sqrt(4+r^2) 8(3r^2)

so

(2r^3)(sqrt(4+r^2)) - ((4/3)(r^3)*(4+r^2)^3/2))

then

[(2/3)(r^3)] * (4+r^2) =

[(10/3)(r^3+r^5)] + C. <<----- this is ur answer,
i think this is better.

Answer 5

((x² - 8)*√(x² + 4))/3 evaluated between 0 and 1 ☺


Doug

Answer 6

/1 r^3
| ------------------ dr
/0 (4+x^2)^(1/2)


= (1/3)(-8 + r^2)(4 + r^2)^(1/2) + C