Question

Evaluate the following.

${\displaystyle \int \frac{20-12{\text{e}}^{\text{x}}}{3{\text{e}}^{\text{x}}-4}}$dx

Answer 1

Let I = ${\displaystyle \int \frac{20-12{\text{e}}^{\text{x}}}{3{\text{e}}^{\text{x}}-4}}$dx

Let 20 - 12e^x = A(3e^x - 4) + B ${\displaystyle \frac{\text{d}}{\text{dx}}}$(3e^x - 4)

= 3 Ae^x - 4A + 3Be^x 

∴ 20 - 12e^x = (3A + 3B)e^x - 4A

Comparing the coefficients of e^x and constant term on both sides, we get

- 4A = 20 and 3A + 3B = - 12

Solving these equations, we get

A = -5 and B = 1

∴ I = ${\displaystyle \int \frac{-5(3{\text{e}}^{\text{x}}-4)+3{\text{e}}^{\text{x}}}{3{\text{e}}^{\text{x}}-4}}$dx

${\displaystyle =-5\int \text{dx}+\int \frac{3{\text{e}}^{\text{x}}}{3{\text{e}}^{\text{x}}-4}}$ dx

∴ I = - 5x + log ${\displaystyle \left|(3{\text{e}}^{\text{x}}-4)\right|}$ + c   ....${\displaystyle [\int \frac{\text{f}\prime \left(\text{x}\right)}{\text{f}\left(\text{x}\right)}\text{dx}=\log \left|f\left(\text{x}\right)\right|+\text{c}]}$