What is the integral of (2e^{2x})/(e^{2x)+16e^x+63} ?

The integral of (2e^{2x})/(e^{2x)+16e^x+63} is ln|16e^x+e^{2x}+63|+8ln|e^x+9|-8ln|e^x+7|+C

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integral of (2e^{2x})/(e^{2x)+16e^x+63}

\int \frac{2 e ^{2 x}}{e ^{2 x} + 16 e ^{x} + 63} d x

ln 16 e^{x} + e^{2 x} + 63 + 8 ln ∣ e^{x} + 9 ∣ - 8 ln ∣ e^{x} + 7 ∣ + C

Solution steps

\int \frac{2 e ^{2 x}}{e ^{2 x} + 16 e ^{x} + 63} 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} + 16 e ^{x} + 63} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 16 u + 63} d u

= 2 \cdot \int \frac{u}{u ^{2} + 16 u + 63} d u

Complete the square

u^{2} + 16 u + 63 : (u + 8)^{2} - 1

= 2 \cdot \int \frac{u}{( u + 8 ) ^{2} - 1} d u

Apply u-substitution

: \int \frac{v - 8}{v ^{2} - 1} d v

= 2 \cdot \int \frac{v - 8}{v ^{2} - 1} d v

Expand

\frac{v - 8}{v ^{2} - 1} : \frac{v}{v ^{2} - 1} - \frac{8}{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{8}{v ^{2} - 1} d v)

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

\int \frac{8}{v ^{2} - 1} d v = - 8 (\frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣)

= 2 (\frac{1}{2} ln v^{2} - 1 - (- 8 (\frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣)))

Substitute back

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

Simplify

2 (\frac{1}{2} ln (e^{x} + 8)^{2} - 1 - (- 8 (\frac{1}{2} ln ∣ e^{x} + 8 + 1 ∣ - \frac{1}{2} ln ∣ e^{x} + 8 - 1 ∣))) : ln 16 e^{x} + e^{2 x} + 63 + 8 ln ∣ e^{x} + 9 ∣ - 8 ln ∣ e^{x} + 7 ∣

= ln 16 e^{x} + e^{2 x} + 63 + 8 ln ∣ e^{x} + 9 ∣ - 8 ln ∣ e^{x} + 7 ∣

Add a constant to the solution

= ln 16 e^{x} + e^{2 x} + 63 + 8 ln ∣ e^{x} + 9 ∣ - 8 ln ∣ e^{x} + 7 ∣ + C

ln (x^{6})

\int_{0}^{1} \frac{x ^{4}}{x ^{5} - 1} d x

n = 0 \sum \infty 2 (- 5)^{n}

\frac{1}{s ^{3} - 7 s}

y^{^{'''}} + 64 y^{^{'}} = 0

What is the integral of (2e^{2x})/(e^{2x)+16e^x+63} ?
The integral of (2e^{2x})/(e^{2x)+16e^x+63} is ln|16e^x+e^{2x}+63|+8ln|e^x+9|-8ln|e^x+7|+C