How to integrate !!!!!!!!!!!!!!!!!!!!!!!!!!!!

Question

(2x+3)/(3x-4) dx?

Answer 1

Apply long division

Integrate the sum term by term

Where u = ( -4+3x)

du =3dx

The solution is  .

Answer 2

∫(2x + 3)/(3x - 4) dx

Let u = 3x - 4   => 3x = u + 4 => x = (u + 4)/3

 => du / dx = 3

 => du/3 = dx

And 2x+3 = 2(u +4)/3 +3

                  = 2u/3 + 8/3 + 3

                  = 2u/3 + (8 + 9)/3

                  = 2u/3 + 17/3

Therefore ∫(2x + 3)/(3x - 4) dx = 1/3* ∫(2u/3 + 17/3)/u  du

                                                   = 1/9 * ∫(2 + 17/u) du

                                                   = 1/9 * ∫2 du + 1/9 * ∫17/u du

                                                   = 2/9 * ∫ du + 17/9 * ∫1/u du

                                                   = 2u/9 + 17/9log|u| + c       [ Since ∫ du = u + c, ∫1/u du = log|u| + c ]

                                                   = 2(3x+4)/9 + 17/9 log|3x - 4| + C       [ Where u = 3x-4 ]

                                                   = 6x/9 + 8/9 + 17/9 log|3x - 4| + C

                                                   = 3x/2 + 17/9ln|3x - 4| + C

Therefore ∫(2x + 3)/(3x - 4) dx = 3x/2 + 17/9ln|3x - 4| + C
 

Comments

6x/9=2x/3 not 3x/2