What is the integral of ((e^x+9))/(e^{2x)-6e^x+9} ?

The integral of ((e^x+9))/(e^{2x)-6e^x+9} is x-4/(e^x-3)-ln|e^x-3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((e^x+9))/(e^{2x)-6e^x+9}

\int \frac{( e ^{x} + 9 )}{e ^{2 x} - 6 e ^{x} + 9} d x

x - \frac{4}{e ^{x} - 3} - ln ∣ e^{x} - 3 ∣ + C

Solution steps

\int \frac{e ^{x} + 9}{e ^{2 x} - 6 e ^{x} + 9} d x

Expand

\frac{e ^{x} + 9}{e ^{2 x} - 6 e ^{x} + 9} : \frac{e ^{x}}{e ^{2 x} - 6 e ^{x} + 9} + \frac{9}{e ^{2 x} - 6 e ^{x} + 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{e ^{x}}{e ^{2 x} - 6 e ^{x} + 9} d x + \int \frac{9}{e ^{2 x} - 6 e ^{x} + 9} d x

\int \frac{e ^{x}}{e ^{2 x} - 6 e ^{x} + 9} d x = - \frac{1}{e ^{x} - 3}

\int \frac{9}{e ^{2 x} - 6 e ^{x} + 9} d x = x - ln ∣ e^{x} - 3 ∣ - \frac{3}{e ^{x} - 3}

= - \frac{1}{e ^{x} - 3} + x - ln ∣ e^{x} - 3 ∣ - \frac{3}{e ^{x} - 3}

Simplify

- \frac{1}{e ^{x} - 3} + x - ln ∣ e^{x} - 3 ∣ - \frac{3}{e ^{x} - 3} : x - \frac{4}{e ^{x} - 3} - ln ∣ e^{x} - 3 ∣

= x - \frac{4}{e ^{x} - 3} - ln ∣ e^{x} - 3 ∣

Add a constant to the solution

= x - \frac{4}{e ^{x} - 3} - ln ∣ e^{x} - 3 ∣ + C

f (x) = x^{2} + 3, (- 3, 12)

\frac{\partial}{\partial x} (f (x, y) (x + y))

y^{^{''}} + y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d}{d x} (\frac{1 - 3 x - x ^{2}}{x ^{2} - 4})

\frac{\partial}{\partial x} (\frac{x}{x ^{5} - y ^{7}})

What is the integral of ((e^x+9))/(e^{2x)-6e^x+9} ?
The integral of ((e^x+9))/(e^{2x)-6e^x+9} is x-4/(e^x-3)-ln|e^x-3|+C