What is the integral of (3e^{2x})/(e^{2x)+13e^x+40} ?

The integral of (3e^{2x})/(e^{2x)+13e^x+40} is

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integral of (3e^{2x})/(e^{2x)+13e^x+40}

\int \frac{3 e ^{2 x}}{e ^{2 x} + 13 e ^{x} + 40} d x

6 (\frac{1}{4} ln 52 e^{x} + 4 e^{2 x} + 160 + \frac{13}{12} (ln \frac{16}{3} + \frac{2}{3} e^{x} - ln \frac{10}{3} + \frac{2}{3} e^{x})) + C

Solution steps

\int \frac{3 e ^{2 x}}{e ^{2 x} + 13 e ^{x} + 40} d x

Take the constant out:

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

= 3 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 13 e ^{x} + 40} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 13 u + 40} d u

= 3 \cdot \int \frac{u}{u ^{2} + 13 u + 40} d u

Complete the square

u^{2} + 13 u + 40 : (u + \frac{13}{2})^{2} - \frac{9}{4}

= 3 \cdot \int \frac{u}{( u + \frac{13}{2} ) ^{2} - \frac{9}{4}} d u

Apply u-substitution

: \int \frac{2 ( 2 v - 13 )}{4 v ^{2} - 9} d v

= 3 \cdot \int \frac{2 ( 2 v - 13 )}{4 v ^{2} - 9} d v

Take the constant out:

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

= 3 \cdot 2 \cdot \int \frac{2 v - 13}{4 v ^{2} - 9} d v

Expand

\frac{2 v - 13}{4 v ^{2} - 9} : \frac{2 v}{4 v ^{2} - 9} - \frac{13}{4 v ^{2} - 9}

Apply the Sum Rule:

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

= 3 \cdot 2 (\int \frac{2 v}{4 v ^{2} - 9} d v - \int \frac{13}{4 v ^{2} - 9} d v)

\int \frac{2 v}{4 v ^{2} - 9} d v = \frac{1}{4} ln 4 v^{2} - 9

\int \frac{13}{4 v ^{2} - 9} d v = - \frac{13}{12} (ln \frac{2}{3} v + 1 - ln \frac{2}{3} v - 1)

= 3 \cdot 2 (\frac{1}{4} ln 4 v^{2} - 9 - (- \frac{13}{12} (ln \frac{2}{3} v + 1 - ln \frac{2}{3} v - 1)))

Substitute back

= 3 \cdot 2 (\frac{1}{4} ln 4 (e^{x} + \frac{13}{2})^{2} - 9 - (- \frac{13}{12} (ln \frac{2}{3} (e^{x} + \frac{13}{2}) + 1 - ln \frac{2}{3} (e^{x} + \frac{13}{2}) - 1)))

Simplify

3 \cdot 2 (\frac{1}{4} ln 4 (e^{x} + \frac{13}{2})^{2} - 9 - (- \frac{13}{12} (ln \frac{2}{3} (e^{x} + \frac{13}{2}) + 1 - ln \frac{2}{3} (e^{x} + \frac{13}{2}) - 1))) : 6 (\frac{1}{4} ln 52 e^{x} + 4 e^{2 x} + 160 + \frac{13}{12} (ln \frac{16}{3} + \frac{2}{3} e^{x} - ln \frac{10}{3} + \frac{2}{3} e^{x}))

= 6 (\frac{1}{4} ln 52 e^{x} + 4 e^{2 x} + 160 + \frac{13}{12} (ln \frac{16}{3} + \frac{2}{3} e^{x} - ln \frac{10}{3} + \frac{2}{3} e^{x}))

Add a constant to the solution

= 6 (\frac{1}{4} ln 52 e^{x} + 4 e^{2 x} + 160 + \frac{13}{12} (ln \frac{16}{3} + \frac{2}{3} e^{x} - ln \frac{10}{3} + \frac{2}{3} e^{x})) + C

x \to 0 lim (- x^{2} + 3 x + a)

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What is the integral of (3e^{2x})/(e^{2x)+13e^x+40} ?
The integral of (3e^{2x})/(e^{2x)+13e^x+40} is