How to do this integration?

Question

(2+3x) / sq root of 4X+3X2

Answer 1

I = ∫ ((2 + 3x) / √3x² + 4x)).dx
let u = 3x² + 4x
du/dx = 6x + 4 = 2.(3x + 2)
du/2 = (3x + 2).dx
I = (1/2) ∫ u^(- 1/2).du
I = (1/2).u^(1/2) / (1/2) + C
I = (3x² + 4x)^(1/2) + C

Answer 2

Integral ( (2 + 3x) / sqrt(4x + 3x^2) dx )

To solve this, you need to use substitution. But before substitution, I'm going to show you an intermediate step to make the substitution obvious. Rearranging the integral, we get

Integral ( 1 / sqrt(4x + 3x^2) (2 + 3x) dx )

Let u = 4x + 3x^2. Then
du = (4 + 6x) dx. Multiply (1/2) both sides, and we get
(1/2) du = (2 + 3x) dx

Note that (2 + 3x) dx is the tail end of our integral, so it follows that (1/2) du is the tail end of our new integral.

Integral ( 1/sqrt(u) (1/2) du )

Factor (1/2) out of the integral,

(1/2) Integral ( 1/sqrt(u) du)

Change sqrt(u) into u^(1/2),

(1/2) Integral ( 1/u^(1/2) du )

Make into a negative exponent.

(1/2) Integral ( u^(-1/2) du )

Use the reverse power rule.

(1/2) (2)u^(3/2) + C

Simplify.

u^(3/2) + C

But u = 4x + 3x^2, so our final answer is

[4x + 3x^2]^(3/2) + C

Answer 3

(2+3x) * (4x + 3x^2)^(-1/2) dx == 1/2 * (4 + 6x) (4x + 3x^2)^(-1/2) dx

Integation yields:
-1/3 * (4x - 3x^2)^(-3/2)

Answer 4

y = (2+3x)/sqrt(3x^2 + 4x)

y' = (3*sqrt(3x^2 + 4x) - (2 + 3x)*0.5*(3x^2 + 4x)^(-0.5)*(6x + 4))/ (3x^2 + 4x)

y' = (3x(3x + 4) - 0.5(3x + 2)(6x + 4))/(3x^2 + 4x)^(1.5) =

= (9x^2 + 12x - 9x^2 - 6x - 6x - 8)/ (3x^2 + 4x)^(1.5) =
= -8/(3x^2 + 4x)^(1.5)