∫(x^3)*(x^2 +4)^(1/2)?

Question

How is this done?

( My first impulse was to set u=x^3, but I don't think that will get the x^2 out of the square-root )

Answer 1

symbol [ stands for integral

[x^3(x^2 + 4)^1/2 dx

x = 2 tanu
dx = 2 sec^2u du

8 [tan^3 u (4(1 + tan^2u)^1/2 (2sec^2u) du

32[ tan^3 u (sec u)(sec^2u) du
32[tan^3 u sec^3 u du
32[tan^2 u sec ^2 u (sec u tan u) du
32[(sec^2 u - 1)sec^2u(sec u tan u) du
32[(sec^2 u - 1)sec^2u d(sec u) (since d(secu = secu tanu)
32[(sec^4 u - sec^2 u d(sec u)
32[ 1/5 sec^5u - 1/3 sec^3 u) + c

since 2tanu = x
tan u = x/2
tan^2 u = x^2/4
sec^2 u - 1= x^2/4
sec^2 u = x^2/4 + 1
sec u = (x^2/4 + 1)^1/2
secu = (1/2)(x^2 + 4)^1/2
substituting this
32((1/5)(1/2)^5(x^2 + 4)^5/2 - (1/3)(1/2)^3(x^2+4)^3/2) + c
1/5(x^2 + 4)^5/2 - (4/3)(x^2 + 4)^3/2 + c

Answer 2

let u = x^2 +4
du = 2x dx
du/2 = xdx

∫x(x^2)(x^2 +4)^(1/2) dx cux x^3 = x^2 *x
∫(x^2)(x^2 +4)^(1/2) xdx (****) i move x infront nearby dx

because we let u = x^2 +4 then x^2 = u-4

replace everything in term of u, u= x^2 +4, x^2 = u-4, xdx = du/2 to equation (****) then you will have
∫(u-4)(u)^(1/2) du/2
1/2 ∫(u-4)(u)^(1/2) du
1/2 ∫u^(3/2)-4u^(1/2) du cux we distibute

1/2 { (2/5)u^(5/2) -4(2/3)u^(3/2)}+C
(1/5)u^(5/2) -(4/3)u^(3/2) +C

replace u = x^2 +4 back

(1/5)(x^2+4)^(5/2) - (4/3)(x^2+4)^(3/2) +C

hope it helps

Answer 3

You can set x = 2sh(u)
dx = 2ch(u) du
sqrt(x² + 4) = 2sqrt(sh²(u) + 1) = 2ch(u)

x³ sqrt(x² + 4)dx = 32 sh³(u)ch²(u) du
= 32 sh(u) (ch²(u) - 1) ch²(u) du
=[32 (ch(u))^4 - 32 ch²(u) ]sh(u) du

the integral is 32/5 (ch(u))^5 - 32/3 ch³(u)

Answer 4

∫ x³ √(x²+ 4) dx =
-----> x = 2 sh(t); t = arsh(x/2)
= ∫ 8sh³(t) √(4sh²(t)+ 4) d [2sh(t)] =
= ∫ 8sh³(t) √(4ch(t)²) 2ch(t) dt =
= 32 ∫ sh³(t) ch²(t) dt =
= 32 ∫ sh²(t) ch²(t) d[ch(t)] =
--------> u = ch(t) = ch(arsh(x/2))
= 32 ∫ (u²-1) u² du =
= 32 [u^5/5 - u^3/3] =


Answer:
32/5 ch^5(arsh(x/2)) - 32/3 ch^3(arsh(x/2)) + C,