What is the integral of (6e^{2x})/(e^{2x)+12e^x+27} ?

The integral of (6e^{2x})/(e^{2x)+12e^x+27} is 6(3/2 ln|e^x+9|-1/2 ln|e^x+3|)+C

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integral of (6e^{2x})/(e^{2x)+12e^x+27}

\int \frac{6 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

6 (\frac{3}{2} ln ∣ e^{x} + 9 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣) + C

Solution steps

\int \frac{6 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

Take the constant out:

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

= 6 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 12 u + 27} d u

= 6 \cdot \int \frac{u}{u ^{2} + 12 u + 27} d u

Complete the square

u^{2} + 12 u + 27 : (u + 6)^{2} - 9

= 6 \cdot \int \frac{u}{( u + 6 ) ^{2} - 9} d u

Apply u-substitution

: \int \frac{v - 6}{v ^{2} - 9} d v

= 6 \cdot \int \frac{v - 6}{v ^{2} - 9} d v

Take the partial fraction of

\frac{v - 6}{v ^{2} - 9} : \frac{3}{2 ( v + 3 )} - \frac{1}{2 ( v - 3 )}

= 6 \cdot \int \frac{3}{2 ( v + 3 )} - \frac{1}{2 ( v - 3 )} 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

= 6 (\int \frac{3}{2 ( v + 3 )} d v - \int \frac{1}{2 ( v - 3 )} d v)

\int \frac{3}{2 ( v + 3 )} d v = \frac{3}{2} ln ∣ v + 3 ∣

\int \frac{1}{2 ( v - 3 )} d v = \frac{1}{2} ln ∣ v - 3 ∣

= 6 (\frac{3}{2} ln ∣ v + 3 ∣ - \frac{1}{2} ln ∣ v - 3 ∣)

Substitute back

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

Simplify

6 (\frac{3}{2} ln ∣ e^{x} + 6 + 3 ∣ - \frac{1}{2} ln ∣ e^{x} + 6 - 3 ∣) : 6 (\frac{3}{2} ln ∣ e^{x} + 9 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣)

= 6 (\frac{3}{2} ln ∣ e^{x} + 9 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣)

Add a constant to the solution

= 6 (\frac{3}{2} ln ∣ e^{x} + 9 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣) + C

\int x (x^{2} + x) d x

2 z (3 z + 1) d w + (1 - 2 w) d z = 0

\frac{\partial}{\partial y} (e^{x y z^{3}} y z^{3})

\int x (a - b x^{2}) d x

f (x) = 2 x^{2} + x - 2, at x = 4

What is the integral of (6e^{2x})/(e^{2x)+12e^x+27} ?
The integral of (6e^{2x})/(e^{2x)+12e^x+27} is 6(3/2 ln|e^x+9|-1/2 ln|e^x+3|)+C