What is the integral of (5e^{2x})/(e^{2x)+11e^x+24} ?

The integral of (5e^{2x})/(e^{2x)+11e^x+24} is

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integral of (5e^{2x})/(e^{2x)+11e^x+24}

\int \frac{5 e ^{2 x}}{e ^{2 x} + 11 e ^{x} + 24} d x

10 (\frac{1}{4} ln 44 e^{x} + 4 e^{2 x} + 96 + \frac{11}{20} (ln \frac{16}{5} + \frac{2}{5} e^{x} - ln \frac{6}{5} + \frac{2}{5} e^{x})) + C

Solution steps

\int \frac{5 e ^{2 x}}{e ^{2 x} + 11 e ^{x} + 24} d x

Take the constant out:

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

= 5 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 11 e ^{x} + 24} d x

Apply u-substitution

= 5 \cdot \int \frac{u}{u ^{2} + 11 u + 24} d u

Complete the square

u^{2} + 11 u + 24 : (u + \frac{11}{2})^{2} - \frac{25}{4}

= 5 \cdot \int \frac{u}{( u + \frac{11}{2} ) ^{2} - \frac{25}{4}} d u

Apply u-substitution

= 5 \cdot \int \frac{2 ( 2 v - 11 )}{4 v ^{2} - 25} d v

Take the constant out:

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

= 5 \cdot 2 \cdot \int \frac{2 v - 11}{4 v ^{2} - 25} d v

Expand

\frac{2 v - 11}{4 v ^{2} - 25} : \frac{2 v}{4 v ^{2} - 25} - \frac{11}{4 v ^{2} - 25}

Apply the Sum Rule:

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

= 5 \cdot 2 (\int \frac{2 v}{4 v ^{2} - 25} d v - \int \frac{11}{4 v ^{2} - 25} d v)

\int \frac{2 v}{4 v ^{2} - 25} d v = \frac{1}{4} ln 4 v^{2} - 25

\int \frac{11}{4 v ^{2} - 25} d v = - \frac{11}{20} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1)

= 5 \cdot 2 (\frac{1}{4} ln 4 v^{2} - 25 - (- \frac{11}{20} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1)))

Substitute back

= 5 \cdot 2 (\frac{1}{4} ln 4 (e^{x} + \frac{11}{2})^{2} - 25 - (- \frac{11}{20} (ln \frac{2}{5} (e^{x} + \frac{11}{2}) + 1 - ln \frac{2}{5} (e^{x} + \frac{11}{2}) - 1)))

Simplify

5 \cdot 2 (\frac{1}{4} ln 4 (e^{x} + \frac{11}{2})^{2} - 25 - (- \frac{11}{20} (ln \frac{2}{5} (e^{x} + \frac{11}{2}) + 1 - ln \frac{2}{5} (e^{x} + \frac{11}{2}) - 1))) : 10 (\frac{1}{4} ln 44 e^{x} + 4 e^{2 x} + 96 + \frac{11}{20} (ln \frac{16}{5} + \frac{2}{5} e^{x} - ln \frac{6}{5} + \frac{2}{5} e^{x}))

= 10 (\frac{1}{4} ln 44 e^{x} + 4 e^{2 x} + 96 + \frac{11}{20} (ln \frac{16}{5} + \frac{2}{5} e^{x} - ln \frac{6}{5} + \frac{2}{5} e^{x}))

Add a constant to the solution

= 10 (\frac{1}{4} ln 44 e^{x} + 4 e^{2 x} + 96 + \frac{11}{20} (ln \frac{16}{5} + \frac{2}{5} e^{x} - ln \frac{6}{5} + \frac{2}{5} e^{x})) + C

What is the integral of (5e^{2x})/(e^{2x)+11e^x+24} ?
The integral of (5e^{2x})/(e^{2x)+11e^x+24} is