What is the integral of (4e^{2x})/(e^{2x)+10e^x+21} ?

The integral of (4e^{2x})/(e^{2x)+10e^x+21} is 7ln|e^x+7|-3ln|e^x+3|+C

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integral of (4e^{2x})/(e^{2x)+10e^x+21}

\int \frac{4 e ^{2 x}}{e ^{2 x} + 10 e ^{x} + 21} d x

7 ln ∣ e^{x} + 7 ∣ - 3 ln ∣ e^{x} + 3 ∣ + C

Solution steps

\int \frac{4 e ^{2 x}}{e ^{2 x} + 10 e ^{x} + 21} d x

Take the constant out:

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

= 4 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 10 e ^{x} + 21} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 10 u + 21} d u

= 4 \cdot \int \frac{u}{u ^{2} + 10 u + 21} d u

Complete the square

u^{2} + 10 u + 21 : (u + 5)^{2} - 4

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

Apply u-substitution

: \int \frac{v - 5}{v ^{2} - 4} d v

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

Take the partial fraction of

\frac{v - 5}{v ^{2} - 4} : \frac{7}{4 ( v + 2 )} - \frac{3}{4 ( v - 2 )}

= 4 \cdot \int \frac{7}{4 ( v + 2 )} - \frac{3}{4 ( v - 2 )} d v

Apply the Sum Rule:

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

= 4 (\int \frac{7}{4 ( v + 2 )} d v - \int \frac{3}{4 ( v - 2 )} d v)

\int \frac{7}{4 ( v + 2 )} d v = \frac{7}{4} ln ∣ v + 2 ∣

\int \frac{3}{4 ( v - 2 )} d v = \frac{3}{4} ln ∣ v - 2 ∣

= 4 (\frac{7}{4} ln ∣ v + 2 ∣ - \frac{3}{4} ln ∣ v - 2 ∣)

Substitute back

= 4 (\frac{7}{4} ln ∣ e^{x} + 5 + 2 ∣ - \frac{3}{4} ln ∣ e^{x} + 5 - 2 ∣)

Simplify

4 (\frac{7}{4} ln ∣ e^{x} + 5 + 2 ∣ - \frac{3}{4} ln ∣ e^{x} + 5 - 2 ∣) : 7 ln ∣ e^{x} + 7 ∣ - 3 ln ∣ e^{x} + 3 ∣

= 7 ln ∣ e^{x} + 7 ∣ - 3 ln ∣ e^{x} + 3 ∣

Add a constant to the solution

= 7 ln ∣ e^{x} + 7 ∣ - 3 ln ∣ e^{x} + 3 ∣ + C

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What is the integral of (4e^{2x})/(e^{2x)+10e^x+21} ?
The integral of (4e^{2x})/(e^{2x)+10e^x+21} is 7ln|e^x+7|-3ln|e^x+3|+C