What is the integral of (e^{2x})/(e^{2x)-4e^x+3} ?

The integral of (e^{2x})/(e^{2x)-4e^x+3} is 1/2 ln|-4e^x+e^{2x}+3|-ln|e^x-1|+ln|e^x-3|+C

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integral of (e^{2x})/(e^{2x)-4e^x+3}

\int \frac{e ^{2 x}}{e ^{2 x} - 4 e ^{x} + 3} d x

\frac{1}{2} ln - 4 e^{x} + e^{2 x} + 3 - ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 3 ∣ + C

Solution steps

\int \frac{e ^{2 x}}{e ^{2 x} - 4 e ^{x} + 3} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 4 u + 3} d u

= \int \frac{u}{u ^{2} - 4 u + 3} d u

Complete the square

u^{2} - 4 u + 3 : (u - 2)^{2} - 1

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

Apply u-substitution

: \int \frac{v + 2}{v ^{2} - 1} d v

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

Expand

\frac{v + 2}{v ^{2} - 1} : \frac{v}{v ^{2} - 1} + \frac{2}{v ^{2} - 1}

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

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

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

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

Substitute back

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

Simplify

\frac{1}{2} ln (e^{x} - 2)^{2} - 1 - ln ∣ e^{x} - 2 + 1 ∣ + ln ∣ e^{x} - 2 - 1 ∣ : \frac{1}{2} ln - 4 e^{x} + e^{2 x} + 3 - ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 3 ∣

= \frac{1}{2} ln - 4 e^{x} + e^{2 x} + 3 - ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 3 ∣

Add a constant to the solution

= \frac{1}{2} ln - 4 e^{x} + e^{2 x} + 3 - ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 3 ∣ + C

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What is the integral of (e^{2x})/(e^{2x)-4e^x+3} ?
The integral of (e^{2x})/(e^{2x)-4e^x+3} is 1/2 ln|-4e^x+e^{2x}+3|-ln|e^x-1|+ln|e^x-3|+C