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

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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) = \frac{8}{x}, at x = 4

y = 2 x, 2 x - 4, y = 0

5 x^{3} - 10 x^{2} - \frac{3}{x}

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

\frac{d y}{d t} = 5 y + y^{4}, y (0) = 1

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