<p>$\int{\frac{1+{{e}^{-x}}}{1-4{{e}^{-x}}}\mathrm{ d}x}$</p>

$\displaystyle\displaystyle \int{\frac{1+{{e}^{-x}}}{1-4{{e}^{-x}}}\mathrm{ d}x}$

Techniques of  Integration - General

Techniques of  Integration

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

ACJC Techniques of Integration Tutorial Q16

**Method I (Split it):**

$\displaystyle\int \frac{1+e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$

$\displaystyle = \displaystyle\int \frac{1}{1-4e^{-x}} + \frac{e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$

$\displaystyle = \displaystyle\int \frac{e^x}{e^x(1-4e^{-x})} \, \mathrm{d}x + \displaystyle\int \frac{e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$; let $u = 1 - 4e^{-x}$

$\displaystyle = \displaystyle\int \frac{e^x}{e^x-4} \, \mathrm{d}x + \displaystyle\int \frac{e^{-x}}{u} \cdot \frac{\mathrm{d}u}{4e^{-x}}$

$\displaystyle = \frac{1}{4}\displaystyle\int \frac{1}{u} \, du = \frac{1}{4}\ln|u|$

$= \ln|e^x - 4| + \frac{1}{4}\ln|1 - 4e^{-x}| + c$

**Method II (Juggle):**

$\displaystyle\int \frac{1+e^{-x}+4e^{-x}-4e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$

$\displaystyle = \displaystyle\int \frac{(1-4e^{-x}) + e^{-x} + 4e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$

$\displaystyle = \displaystyle\int \frac{1-4e^{-x} + 5e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$

$\displaystyle = \displaystyle\int \frac{1-4e^{-x}}{1-4e^{-x}} + \frac{5e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$

$\displaystyle = \displaystyle\int 1 \, \mathrm{d}x + \frac{5}{4}\displaystyle\int \frac{4e^{-x}}{1-4e^{-x}} \, \mathrm{d}x$

$\displaystyle = x + \frac{5}{4}\ln|1-4e^{-x}| + c$

ACJC Techniques of Integration Tutorial Q16

ACJC Techniques of Integration Tutorial Q16

Video Solution: