Verified answer

1) ∫ √(4x³) dx = 2∫ x^(3/2) dx

= 2 * 2/5 * x^(5/2)

= (4/5)x^(5/2) + C

====================

EDIT:

to explain more clearly, √(4x³) = 2√(x³)

so ∫ √(4x³) dx = 2∫ √(x³) dx = 2∫ x^(3/2) dx

let me know if you need more info

====================

2) ∫ √(4x² + 4) dx

= 2∫ √(x² + 1) dx

Now this is a circle and a B**tch to integrate

= x√(x² + 1) + arcsinh(x) + C

Got this from online integrator - link given below

or

let x = tanθ, hence secθ = √(x² + 1)

dx = sec²θdθ

2∫ √(x² + 1) dx = 2∫ √(tan²θ + 1) sec²θdθ

= 2∫ √(sec²θ) sec²θdθ

= 2∫ sec³θdθ

let u = secθ, dv = sec²θdθ

du = secθtanθdθ, v = tanθ

= 2∫sec³θdθ

= 2(secθtanθ - ∫secθtan²θdθ)

= 2(secθtanθ - ∫secθ(sec²θ - 1)dθ)

= 2(secθtanθ - ∫sec³θ - secθdθ)

= 2(secθtanθ - ∫sec³θdθ + ∫secθdθ)

= 2(secθtanθ - ∫sec³θdθ + ∫secθ((secθ + tanθ)(secθ + tanθ))dθ)

= 2(secθtanθ - ∫sec³θdθ + ∫(sec²θ + secθtanθ)(secθ + tanθ)dθ)

u = secθ + tanθ

du = secθtanθ + sec²θ (so numerator is derivative of denominator)

= 2(secθtanθ - ∫sec³θdθ + ln|secθ + tanθ|)

so

2∫sec³θdθ = 2(secθtanθ - ∫sec³θdθ + ln|secθ + tanθ|)

∫sec³θdθ = secθtanθ - ∫sec³θdθ + ln|secθ + tanθ|

2∫sec³θdθ = secθtanθ + ln|secθ + tanθ|

2∫ √(x² + 1) dx = 2∫sec³θdθ = secθtanθ + ln|secθ + tanθ| + C

This is the solution but in terms of θ, need it in terms of x

Note x = tanθ, hence secθ = √(x² + 1)

2∫ √(x² + 1) dx = secθtanθ + ln|secθ + tanθ| + C

= x√(x² + 1) + ln | x + √(x² + 1) | + C

3) ∫ √(x² - 1) dx (also a nasty one! but given in my table of standard integrals)

= ½(x√(x² - 1) - ln|x + √(x² - 1)| + C

or:

let x = secθ, so tanθ = √(x² - 1)

dx = secθtanθdθ

∫ √(x² - 1) dx

= ∫√(sec²θ - 1) secθtanθdθ

= ∫tanθsecθtanθdθ

= ∫secθtan²θdθ

let u = tanθ dv = secθtanθdθ

du = sec²θdθ, v = secθ

= secθtanθ - ∫secθsec²θdθ

= secθtanθ - ∫sec³θdθ (note this is already solved above)

∫sec³θdθ = ½ secθtanθ + ½ ln|secθ + tanθ|

so

∫ √(x² - 1) dx = secθtanθ - ½ secθtanθ - ½ ln|secθ + tanθ| + C

= ½ secθtanθ - ½ ln|secθ + tanθ| + C

= ½ x√(x² - 1) - ½ ln|x + √(x² - 1)| + C

remembering that x = secθ, so tanθ = √(x² - 1)

1. 1) ∫ √(4x^3) dx

Extremely simple. Just treat it as (4x^3)^(1/2), and bring each term to the power of (1/2). This yields

∫ (4x^3)^(1/2) dx

∫ ( 2x^(3/2) dx )

2 ∫ (x^(3/2) dx )

And now, use the reverse power rule.

2 (2/5)x^(5/2) + C

(4/5)x^(5/2) + C

2. ∫ ( √(4x^2 + 4) dx )

Factoring out the 4 inside the root means it comes out of the root (and the integral) as a 2.

2 ∫ ( √(x^2 + 1) dx )

Use trigonometric substitution to solve.

Let x = tan(t). Then

dx = sec^2(t) dt

2 ∫ ( √(tan^2(t) + 1) sec^2(t) dt )

∫ ( √(sec^2(t)) sec^2(t) dt )

∫ ( sec(t) sec^2(t) dt )

∫ ( sec^3(t) dt )

Which I'll let you solve on your own; you have to use parts and use the fact that it loops around itself.