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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Complete the square

u + u^{2} - 2 : (u + \frac{1}{2})^{2} - \frac{9}{4}

= \int \frac{1}{( u + \frac{1}{2} ) ^{2} - \frac{9}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} - 9} d v

= \int \frac{4}{4 v ^{2} - 9} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{4 v ^{2} - 9} d v

Apply Integral Substitution

= 4 \cdot \int \frac{1}{6 ( w ^{2} - 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

w^{2} - 1 : - (- w^{2} + 1)

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{6} (- \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

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

= 4 \cdot \frac{1}{6} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( e ^{x} + \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( e ^{x} + \frac{1}{2} ) - 1}{2}))

Simplify

4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( e ^{x} + \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( e ^{x} + \frac{1}{2} ) - 1}{2})) : - \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} e^{x} - ln - \frac{2}{3} + \frac{2}{3} e^{x})

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2} + 8)

(x + y) d x + x ln (x) d y = 0

\frac{\partial}{\partial y} (x e^{x y})

x \to 7 lim (tan (\frac{π x}{6}))

f (x) = 3 + 4 x^{2} - 2 x^{3}, (1, 5)

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