What is the integral of (e^x+9)/((e^x-3)^2) ?

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

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integral of (e^x+9)/((e^x-3)^2)

\int \frac{e ^{x} + 9}{( e ^{x} - 3 ) ^{2}} d x

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

Solution steps

\int \frac{e ^{x} + 9}{( e ^{x} - 3 ) ^{2}} d x

e^{x} - 3 = (e^{\frac{x}{2}} + 3) (e^{\frac{x}{2}} - 3)

= \int \frac{e ^{x} + 9}{( ( e ^{\frac{x}{2}} + 3 ) ( e ^{\frac{x}{2}} - 3 ) ) ^{2}} d x

Simplify

\frac{e ^{x} + 9}{( ( e ^{\frac{x}{2}} + 3 ) ( e ^{\frac{x}{2}} - 3 ) ) ^{2}} : \frac{e ^{x} + 9}{- 6 e ^{x} + e ^{2 x} + 9}

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

Expand

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

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

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

\frac{\partial}{\partial x} (\frac{v ^{2}}{x})

\int - ln (x) e^{- x} d x

f (x) = 4 x e^{x}, at x = 0

\int \frac{x ^{5} - x ^{4} + 4 x ^{3} - 4 x ^{2} + 8 x - 4}{( x ^{2} + 2 ) ^{3}} d x

- \frac{6 x}{( x ^{2} - 4 ) ^{2}}

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