What is the integral of 1/((e^x-3)^2) ?

The integral of 1/((e^x-3)^2) is 1/9 x-1/9 ln|e^x-3|-1/(3(e^x-3))+C

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integral of 1/((e^x-3)^2)

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

\frac{1}{9} x - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )} + C

Solution steps

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

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

= \int \frac{1}{( ( e ^{\frac{x}{2}} + 3 ) ( e ^{\frac{x}{2}} - 3 ) ) ^{2}} d x

Simplify

\frac{1}{( ( e ^{\frac{x}{2}} + 3 ) ( e ^{\frac{x}{2}} - 3 ) ) ^{2}} : \frac{1}{- 6 e ^{x} + e ^{2 x} + 9}

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

Apply u-substitution

: \int \frac{1}{u ( - 6 u + u ^{2} + 9 )} d u

= \int \frac{1}{u ( - 6 u + u ^{2} + 9 )} d u

Take the partial fraction of

\frac{1}{u ( - 6 u + u ^{2} + 9 )} : \frac{1}{9 u} - \frac{1}{9 ( u - 3 )} + \frac{1}{3 ( u - 3 ) ^{2}}

= \int \frac{1}{9 u} - \frac{1}{9 ( u - 3 )} + \frac{1}{3 ( u - 3 ) ^{2}} d u

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{1}{9 u} d u - \int \frac{1}{9 ( u - 3 )} d u + \int \frac{1}{3 ( u - 3 ) ^{2}} d u

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u - 3 )} d u = \frac{1}{9} ln ∣ u - 3 ∣

\int \frac{1}{3 ( u - 3 ) ^{2}} d u = - \frac{1}{3 ( u - 3 )}

= \frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u - 3 ∣ - \frac{1}{3 ( u - 3 )}

Substitute back

u = e^{x}

= \frac{1}{9} ln ∣ e^{x} ∣ - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )}

\frac{1}{9} ln ∣ e^{x} ∣ = \frac{1}{9} x

= \frac{1}{9} x - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )}

Add a constant to the solution

= \frac{1}{9} x - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )} + C

x \to 64 lim (x^{\frac{2}{3}})

e^{- t} \cdot sin (t)

\int x d (ln (x)) d x

\int \frac{x}{5 + x} d x

\frac{d}{d x} (\frac{( 5 x - 3 )}{4 x - \frac{3}{x}})

What is the integral of 1/((e^x-3)^2) ?
The integral of 1/((e^x-3)^2) is 1/9 x-1/9 ln|e^x-3|-1/(3(e^x-3))+C