What is the integral of 1/(e^{2x)-4e^x+4} ?

The integral of 1/(e^{2x)-4e^x+4} is x/4-1/4 ln|e^x-2|-1/(2(e^x-2))+C

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integral of 1/(e^{2x)-4e^x+4}

\int \frac{1}{e ^{2 x} - 4 e ^{x} + 4} d x

\frac{x}{4} - \frac{1}{4} ln ∣ e^{x} - 2 ∣ - \frac{1}{2 ( e ^{x} - 2 )} + C

Solution steps

\int \frac{1}{e ^{2 x} - 4 e ^{x} + 4} d x

Apply u-substitution

= \int \frac{1}{u ( u ^{2} - 4 u + 4 )} d u

Complete the square

u^{2} - 4 u + 4 : (u - 2)^{2}

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

Apply u-substitution

= \int \frac{1}{( v + 2 ) v ^{2}} d v

Take the partial fraction of

\frac{1}{( v + 2 ) v ^{2}} : \frac{1}{4 ( v + 2 )} - \frac{1}{4 v} + \frac{1}{2 v ^{2}}

= \int \frac{1}{4 ( v + 2 )} - \frac{1}{4 v} + \frac{1}{2 v ^{2}} d v

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}{4 ( v + 2 )} d v - \int \frac{1}{4 v} d v + \int \frac{1}{2 v ^{2}} d v

\int \frac{1}{4 ( v + 2 )} d v = \frac{1}{4} ln ∣ v + 2 ∣

\int \frac{1}{4 v} d v = \frac{1}{4} ln ∣ v ∣

\int \frac{1}{2 v ^{2}} d v = - \frac{1}{2 v}

= \frac{1}{4} ln ∣ v + 2 ∣ - \frac{1}{4} ln ∣ v ∣ - \frac{1}{2 v}

Substitute back

= \frac{1}{4} ln ∣ e^{x} - 2 + 2 ∣ - \frac{1}{4} ln ∣ e^{x} - 2 ∣ - \frac{1}{2 ( e ^{x} - 2 )}

\frac{1}{4} ln ∣ e^{x} - 2 + 2 ∣ = \frac{x}{4}

= \frac{x}{4} - \frac{1}{4} ln ∣ e^{x} - 2 ∣ - \frac{1}{2 ( e ^{x} - 2 )}

Add a constant to the solution

= \frac{x}{4} - \frac{1}{4} ln ∣ e^{x} - 2 ∣ - \frac{1}{2 ( e ^{x} - 2 )} + C

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What is the integral of 1/(e^{2x)-4e^x+4} ?
The integral of 1/(e^{2x)-4e^x+4} is x/4-1/4 ln|e^x-2|-1/(2(e^x-2))+C