What is the integral of (3e^x)/(e^{2x)+9e^x+20} ?

The integral of (3e^x)/(e^{2x)+9e^x+20} is 3(ln|e^x+4|-ln|e^x+5|)+C

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integral of (3e^x)/(e^{2x)+9e^x+20}

\int \frac{3 e ^{x}}{e ^{2 x} + 9 e ^{x} + 20} d x

3 (ln ∣ e^{x} + 4 ∣ - ln ∣ e^{x} + 5 ∣) + C

Solution steps

\int \frac{3 e ^{x}}{e ^{2 x} + 9 e ^{x} + 20} 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 ^{x}}{e ^{2 x} + 9 e ^{x} + 20} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9 u + 20} d u

= 3 \cdot \int \frac{1}{u ^{2} + 9 u + 20} d u

Take the partial fraction of

\frac{1}{u ^{2} + 9 u + 20} : \frac{1}{u + 4} - \frac{1}{u + 5}

= 3 \cdot \int \frac{1}{u + 4} - \frac{1}{u + 5} d u

Apply the Sum Rule:

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

= 3 (\int \frac{1}{u + 4} d u - \int \frac{1}{u + 5} d u)

\int \frac{1}{u + 4} d u = ln ∣ u + 4 ∣

\int \frac{1}{u + 5} d u = ln ∣ u + 5 ∣

= 3 (ln ∣ u + 4 ∣ - ln ∣ u + 5 ∣)

Substitute back

u = e^{x}

= 3 (ln ∣ e^{x} + 4 ∣ - ln ∣ e^{x} + 5 ∣)

Add a constant to the solution

= 3 (ln ∣ e^{x} + 4 ∣ - ln ∣ e^{x} + 5 ∣) + C

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What is the integral of (3e^x)/(e^{2x)+9e^x+20} ?
The integral of (3e^x)/(e^{2x)+9e^x+20} is 3(ln|e^x+4|-ln|e^x+5|)+C