What is the integral of (4e^{2x})/(e^{2x)+12e^x+32} ?

The integral of (4e^{2x})/(e^{2x)+12e^x+32} is 4(2ln|e^x+8|-ln|e^x+4|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4e^{2x})/(e^{2x)+12e^x+32}

\int \frac{4 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 32} d x

4 (2 ln ∣ e^{x} + 8 ∣ - ln ∣ e^{x} + 4 ∣) + C

Solution steps

\int \frac{4 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 32} d x

Take the constant out:

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

= 4 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 32} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 12 u + 32} d u

= 4 \cdot \int \frac{u}{u ^{2} + 12 u + 32} d u

Complete the square

u^{2} + 12 u + 32 : (u + 6)^{2} - 4

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

Apply u-substitution

: \int \frac{v - 6}{v ^{2} - 4} d v

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

Take the partial fraction of

\frac{v - 6}{v ^{2} - 4} : \frac{2}{v + 2} - \frac{1}{v - 2}

= 4 \cdot \int \frac{2}{v + 2} - \frac{1}{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

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

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

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

= 4 (2 ln ∣ v + 2 ∣ - ln ∣ v - 2 ∣)

Substitute back

= 4 (2 ln ∣ e^{x} + 6 + 2 ∣ - ln ∣ e^{x} + 6 - 2 ∣)

Simplify

= 4 (2 ln ∣ e^{x} + 8 ∣ - ln ∣ e^{x} + 4 ∣)

Add a constant to the solution

= 4 (2 ln ∣ e^{x} + 8 ∣ - ln ∣ e^{x} + 4 ∣) + C

\int \frac{2 x ^{2} - 3 x + 4}{2 x - 7} d x

\int \frac{x ^{6} - x ^{3} + 1}{x ^{4} + 9 x ^{2}} d x

x \to \infty lim (\frac{lo g _{4} ( x )}{lo g _{9} ( x )})

\int_{1}^{5} x - 1 d x

\int \frac{3}{7} x^{- \frac{1}{7}} d x

What is the integral of (4e^{2x})/(e^{2x)+12e^x+32} ?
The integral of (4e^{2x})/(e^{2x)+12e^x+32} is 4(2ln|e^x+8|-ln|e^x+4|)+C