Question

What is `int x^3 * sqrt(x^2 + 4)dx`?

Answer 1

Use the substitution `u = x^2+4` to find

`intx^3sqrt(x^2+4)dx= 1/5(x^2+4)^(5/2) - 4/3(x^2+4)^(3/2)+C`

Explanation:

For this problem we will use:

• `intx^ndx = x^(n+1)/(n+1) + C`

• `int(f(x) + g(x))dx = intf(x)dx + intg(x)dx`

• Integration by Substitution

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We will proceed using the substitution `u = x^2 + 4`.
Then `du = 2xdx` and we have

`intx^3sqrt(x^2+4)dx = 1/2int((x^2+4)-4)sqrt(x^2+4)*2xdx`

`= 1/2int(u-4)sqrt(u)du`

`=1/2int(u^(3/2)-4u^(1/2))du`

`=1/2intu^(3/2)du - 2intu^(1/2)du`

`=1/5u^(5/2) - 4/3u^(3/2)+C`

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

Answer 2

Use `u` substitution and then solve.

Explanation:

`int x^3 * sqrt(x^2 + 4) dx`

Let `u = x^2 + 4`

Differentiate with respect to `x`

`du = 2x dx`
`(du)/2 = x dx`

Also we have `u=x^2+4` therefore `x^2 = u-4`

`int x^3 * sqrt(x^2 + 4) dx`

Rewriting our integral as
`int x^2 * sqrt(x^2 + 4) x dx`

Substituting for `x^2, x^2+4` and `x dx`

`int (u-4) sqrt(u) (du)/2`
`1/2 int (u-4) sqrt(u) du`
`1/2 int (u^(3/2) - 4u^(1/2)) du` Note `sqrt(u) = u^(1/2)`
`1/2 { int u^(3/2) du - 4 int u^(1/2) du }`

We apply power rule `int x^n dx = x^(n+1)/(n+1) + C` where `x` not equal to `1`

`1/2 { u^(3/2 + 1)/(3/2+1) - 4 u^(1/2 + 1)/(1/2+1)} + C`
`1/2{ u^(5/2)/(5/2) - 4 u^(3/2)/(3/2)} + C`
`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`
`1/5 (x^2+4)^(5/2) - 4/3 (x^2+4)^(3/2) +C` Answer .

It can be further simplified if needed.