What is the integral of (2e^{2x})/(e^{2x)+8e^x+15} ?

The integral of (2e^{2x})/(e^{2x)+8e^x+15} is ln|8e^x+e^{2x}+15|+4ln|e^x+5|-4ln|e^x+3|+C

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integral of (2e^{2x})/(e^{2x)+8e^x+15}

\int \frac{2 e ^{2 x}}{e ^{2 x} + 8 e ^{x} + 15} d x

ln 8 e^{x} + e^{2 x} + 15 + 4 ln ∣ e^{x} + 5 ∣ - 4 ln ∣ e^{x} + 3 ∣ + C

Solution steps

\int \frac{2 e ^{2 x}}{e ^{2 x} + 8 e ^{x} + 15} d x

Take the constant out:

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

= 2 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 8 e ^{x} + 15} d x

Apply u-substitution

= 2 \cdot \int \frac{u}{u ^{2} + 8 u + 15} d u

Complete the square

u^{2} + 8 u + 15 : (u + 4)^{2} - 1

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

Apply u-substitution

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= 2 (\frac{1}{2} ln v^{2} - 1 - (- 4 (\frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣)))

Substitute back

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

Simplify

2 (\frac{1}{2} ln (e^{x} + 4)^{2} - 1 - (- 4 (\frac{1}{2} ln ∣ e^{x} + 4 + 1 ∣ - \frac{1}{2} ln ∣ e^{x} + 4 - 1 ∣))) : ln 8 e^{x} + e^{2 x} + 15 + 4 ln ∣ e^{x} + 5 ∣ - 4 ln ∣ e^{x} + 3 ∣

= ln 8 e^{x} + e^{2 x} + 15 + 4 ln ∣ e^{x} + 5 ∣ - 4 ln ∣ e^{x} + 3 ∣

Add a constant to the solution

= ln 8 e^{x} + e^{2 x} + 15 + 4 ln ∣ e^{x} + 5 ∣ - 4 ln ∣ e^{x} + 3 ∣ + C

What is the integral of (2e^{2x})/(e^{2x)+8e^x+15} ?
The integral of (2e^{2x})/(e^{2x)+8e^x+15} is ln|8e^x+e^{2x}+15|+4ln|e^x+5|-4ln|e^x+3|+C