What is the integral of (6e^{2x})/(e^{2x)+10e^x+16} ?

The integral of (6e^{2x})/(e^{2x)+10e^x+16} is 6(4/3 ln|e^x+8|-1/3 ln|e^x+2|)+C

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integral of (6e^{2x})/(e^{2x)+10e^x+16}

\int \frac{6 e ^{2 x}}{e ^{2 x} + 10 e ^{x} + 16} d x

6 (\frac{4}{3} ln ∣ e^{x} + 8 ∣ - \frac{1}{3} ln ∣ e^{x} + 2 ∣) + C

Solution steps

\int \frac{6 e ^{2 x}}{e ^{2 x} + 10 e ^{x} + 16} 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} + 10 e ^{x} + 16} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 10 u + 16} d u

= 6 \cdot \int \frac{u}{u ^{2} + 10 u + 16} d u

Complete the square

u^{2} + 10 u + 16 : (u + 5)^{2} - 9

= 6 \cdot \int \frac{u}{( u + 5 ) ^{2} - 9} d u

Apply u-substitution

: \int \frac{v - 5}{v ^{2} - 9} d v

= 6 \cdot \int \frac{v - 5}{v ^{2} - 9} d v

Take the partial fraction of

\frac{v - 5}{v ^{2} - 9} : \frac{4}{3 ( v + 3 )} - \frac{1}{3 ( v - 3 )}

= 6 \cdot \int \frac{4}{3 ( v + 3 )} - \frac{1}{3 ( 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{4}{3 ( v + 3 )} d v - \int \frac{1}{3 ( v - 3 )} d v)

\int \frac{4}{3 ( v + 3 )} d v = \frac{4}{3} ln ∣ v + 3 ∣

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

= 6 (\frac{4}{3} ln ∣ v + 3 ∣ - \frac{1}{3} ln ∣ v - 3 ∣)

Substitute back

= 6 (\frac{4}{3} ln ∣ e^{x} + 5 + 3 ∣ - \frac{1}{3} ln ∣ e^{x} + 5 - 3 ∣)

Simplify

6 (\frac{4}{3} ln ∣ e^{x} + 5 + 3 ∣ - \frac{1}{3} ln ∣ e^{x} + 5 - 3 ∣) : 6 (\frac{4}{3} ln ∣ e^{x} + 8 ∣ - \frac{1}{3} ln ∣ e^{x} + 2 ∣)

= 6 (\frac{4}{3} ln ∣ e^{x} + 8 ∣ - \frac{1}{3} ln ∣ e^{x} + 2 ∣)

Add a constant to the solution

= 6 (\frac{4}{3} ln ∣ e^{x} + 8 ∣ - \frac{1}{3} ln ∣ e^{x} + 2 ∣) + C

(1 - x) e^{- x}

f (x) = 5 e^{x} - x^{2}

\frac{d}{d x} (x^{3} - 7 x)

\frac{1}{( s ^{2} + 9 )}

\int \frac{x ^{4} - 3 x ^{2} + 5}{x ^{4}} d x

What is the integral of (6e^{2x})/(e^{2x)+10e^x+16} ?
The integral of (6e^{2x})/(e^{2x)+10e^x+16} is 6(4/3 ln|e^x+8|-1/3 ln|e^x+2|)+C