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

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

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

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

- ln ∣ 2 e^{x} + 4 ∣ + ln ∣ 2 e^{x} + 2 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 3 u + 2} d u

= \int \frac{1}{u ^{2} + 3 u + 2} d u

Complete the square

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

= \int \frac{1}{( u + \frac{3}{2} ) ^{2} - \frac{1}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} - 1} d v

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

Factor

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

= - ln ∣ 2 e^{x} + 4 ∣ + ln ∣ 2 e^{x} + 2 ∣ + C

\int 5 e^{5 x} d x

f (x) = - 6 x^{2} - 9, at x = 0

x \to - 4 - lim (f (x))

\frac{d}{d x} (e^{x} \cdot 2^{x})

x \to 0 lim (\frac{e ^{x} - 1}{sin ( 4 x )})

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