Question

How do you use substitution to integrate `x^2sqrt(4x+9)dx`?

Answer 1

The answer can be written as `int x^2sqrt(4x+9)\ dx=\frac{10x^2-18x+27}{140}(4x+9)^{3/2}+C`

Explanation:

Use the substitution `u=4x+9` so that `du=4dx` and `x=(u-9)/4=1/4 u-9/4`. Then

`int x^2sqrt(4x+9)\ dx=1/4 int ((u-9)/4)^2sqrt(u)\ du`

`=1/64 int (u^2-18u+81)u^{1/2}\ du`

`=1/64 int (u^{5/2}-18u^{3/2}+81u^{1/2})\ du`

`=1/64*2/7 u^{7/2}-9/32*2/5 u^{5/2}+81/64*2/3 u^{3/2}+C`

`=(1/224 u^2-9/80 u+27/32) u^{3/2}+C`

Since `u^2=(4x+9)^2=16x^2+72x+81`, we can write this as

`=(1/14 x^2+9/28x+81/224-9/20x-81/80+27/32)(4x+9)^{3/2}+C`

`=\frac{10x^2-18x+27}{140}(4x+9)^{3/2}+C`

Answer 2

You can rewrite this carefully.

`= intx^2sqrt((2sqrtx)^2 + 3^2)dx`

Now it looks like `sqrt(a^2 + x^2)`, which resembles `sqrt(1 + tan^2theta)`.

Let:
`2sqrtx = 3tantheta`
`x = (9tan^2theta)/4`
`x^2 = (81tan^4theta)/16`
`sqrt((2sqrtx)^2 + 3^2) = sqrt(9tan^2theta + 9) = 3sectheta`
`dx = 9/4*2tanthetasec^2thetad theta = 9/2tanthetasec^2thetad theta`

`=> int(81tan^4theta)/16 3sectheta 9/2tanthetasec^2thetad theta`

`= 2187/32inttan^5thetasec^3thetad theta`

Notice how that constant is irreducible... Ugh. It looks a bit crazy, but it's just large numbers. Using some identities:
`= 2187/32int(sec^2theta - 1)^2sec^2thetasecthetatanthetad theta`

And some u-substitution. Let:
`u = sectheta`
`du = secthetatanthetad theta`

`= 2187/32int(u^2 - 1)^2u^2du`

`= 2187/32int(u^4 - 2u^2 + 1)u^2du`

`= 2187/32intu^6 - 2u^4 + u^2du`

`= 2187/32 (u^7/7 - 2/5u^5 + u^3/3)`

`= 2187/32 (sec^7theta/7 - 2/5sec^5theta + sec^3theta/3)`

Draw a right triangle if you want. Since `tantheta = (2sqrtx)/3`, and `sqrt((2sqrtx)^2 + 3^2) = sqrt(4x + 9)`, `sectheta = sqrt(4x+9)/3`:

`= 2187/32 ((4x+9)^(7/2)/(7*3^7) - 2/(5*3^5)(4x+9)^(5/2) + (4x+9)^(3/2)/(3*3^3))`

`= 2187/32 ((4x+9)^(7/2)/(15309) - 2/(1215)(4x+9)^(5/2) + (4x+9)^(3/2)/(81))`

`= 1/32 ((4x+9)^(7/2)/(7) - 18/(5)(4x+9)^(5/2) + 27(4x+9)^(3/2))`

`= 1/32(4x+9)^(3/2)[(4x+9)^(2)/7 - 18/5(4x+9) + 27] + C`

Now let's figure out how to make this look nicer. Get some common denominators, subtract in the numerator, and factor out `8/35`:
`= 1/32(4x+9)^(3/2)[(16x^2 + 72x + 81)/7 - 72/5x - 162/5 + 27]`

`= 1/32(4x+9)^(3/2)[(80x^2 + 360x + 405 - 504x - 1134 + 945)/35]`

`= 1/32(4x+9)^(3/2)[(80x^2 - 144x + 216)/35]`

`= (1*8)/(32*35)(4x+9)^(3/2)[10x^2 - 18x + 27]`

`= 1/140(4x+9)^("3/2")[10x^2 - 18x + 27] + C`