Integration problems?

Question

1.) (2x+3)dx over 3x^2+9x-4

2.) 2sin8xdx over 1- 2cos^2 4x(pls show how on number 2)

3.) (1-4x)dx over 3x-2 (pls show how on number 3


4.) dx over 3^y +3


thank you

Answer 1

1. Integral([(2x+3) / 3x^2+9x-4]dx) To solve this integral you have to do a substitution. Let u = 3x^2 + 9x - 4
the du = 6x + 9 dx factor out a 3 to get
du = 3(2x + 3) dx now divde both sides by 3 to get
du/3 = 2x + 3 dx. Now our integral becomes...
integral([du/3][1/u] you can pull out the 1/3 which gives us
1/3 integral(1/u)du now this is...
1/3[ln u] + c but we have to sub in our value for u which gives
1/3 [ln 3x^2 + 9x -4] + C that is ur final answer.

Answer 2

1- to integrate (2x+3)dx /( 3x^2+9x-4) by subtittuation
let u=3x^2+9x-4
du = 6x +9 dx
du= 3(2x+3) dx ==> dx =du/(3(2x+3)) subtitute this value into the int

int 1/3 (1 /u )du
= 1/3 In u +c =1/3 In (3x^2+9x-4)

2) 2sin8xdx /( 1- 2cos^2 4x)
you know that cos2x = 2cos²x - 1 right
so cos8x =2cos²4x - 1
-cos8x = 1- 2cos²4x right
so int 2sin8x dx /(-cos8x)= -2int (sin8x/cos8x)
let u = cos8x
du = -8sin8x dx ===>dx = du/(-8sin8x)
=-2 int (-1du /(8u)) =int 2/8 int (1/u)
1/4 In u +c
1/4 In cos8x +c

3) try to divide 1-4x by (3x-2 )

then you can get the final solution
1/9 (-12x -5 In (3x-2)+8 )
4) dx over 3^y +3
Are you sure the power to y or x
it suppose to be int dx / 3^x +3

Answer 3

Let u=3x^2+9x-4
du=(6x+9)dx
=3(2x+3)dx

(2x+3)dx=du/3

integral( du/3u)
=1/3*lnu
=1/3 ln(3x^2+9x-4)+ C

Answer 4

crucial through aspects is: ?udv=uv-?vdu so after substitution and putting u=x and dv=F'(x), ?xF'(x)dx=xF(x)-?F(x) and considering that ?F(x)=G(x) because F(x)=G'(x) ?xF'(x)dx=xF(x)-G(x).