What is the integral of (2e^{2x})/(e^{2x)+14e^x+48} ?

The integral of (2e^{2x})/(e^{2x)+14e^x+48} is ln|14e^x+e^{2x}+48|+7ln|e^x+8|-7ln|e^x+6|+C

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integral of (2e^{2x})/(e^{2x)+14e^x+48}

\int \frac{2 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 48} d x

ln 14 e^{x} + e^{2 x} + 48 + 7 ln ∣ e^{x} + 8 ∣ - 7 ln ∣ e^{x} + 6 ∣ + C

Solution steps

\int \frac{2 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 48} 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} + 14 e ^{x} + 48} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 14 u + 48} d u

= 2 \cdot \int \frac{u}{u ^{2} + 14 u + 48} d u

Complete the square

u^{2} + 14 u + 48 : (u + 7)^{2} - 1

= 2 \cdot \int \frac{u}{( u + 7 ) ^{2} - 1} d u

Apply u-substitution

: \int \frac{v - 7}{v ^{2} - 1} d v

= 2 \cdot \int \frac{v - 7}{v ^{2} - 1} d v

Expand

\frac{v - 7}{v ^{2} - 1} : \frac{v}{v ^{2} - 1} - \frac{7}{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{7}{v ^{2} - 1} d v)

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

\int \frac{7}{v ^{2} - 1} d v = - 7 (\frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣)

= 2 (\frac{1}{2} ln v^{2} - 1 - (- 7 (\frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣)))

Substitute back

= 2 (\frac{1}{2} ln (e^{x} + 7)^{2} - 1 - (- 7 (\frac{1}{2} ln ∣ e^{x} + 7 + 1 ∣ - \frac{1}{2} ln ∣ e^{x} + 7 - 1 ∣)))

Simplify

2 (\frac{1}{2} ln (e^{x} + 7)^{2} - 1 - (- 7 (\frac{1}{2} ln ∣ e^{x} + 7 + 1 ∣ - \frac{1}{2} ln ∣ e^{x} + 7 - 1 ∣))) : ln 14 e^{x} + e^{2 x} + 48 + 7 ln ∣ e^{x} + 8 ∣ - 7 ln ∣ e^{x} + 6 ∣

= ln 14 e^{x} + e^{2 x} + 48 + 7 ln ∣ e^{x} + 8 ∣ - 7 ln ∣ e^{x} + 6 ∣

Add a constant to the solution

= ln 14 e^{x} + e^{2 x} + 48 + 7 ln ∣ e^{x} + 8 ∣ - 7 ln ∣ e^{x} + 6 ∣ + C

\int t^{3} 9 t^{4} + 1 d t

\frac{d y}{d x} = 3 x^{2} (1 + y^{2})

y^{^{'}} = \frac{4 x sec ( \frac{y}{x} ) + y}{x}

y^{^{''}} + π y = 0

\int (cos^{2} (t)) d t

What is the integral of (2e^{2x})/(e^{2x)+14e^x+48} ?
The integral of (2e^{2x})/(e^{2x)+14e^x+48} is ln|14e^x+e^{2x}+48|+7ln|e^x+8|-7ln|e^x+6|+C