What is the integral of (4e^{2x})/(e^{2x)+14e^x+45} ?

The integral of (4e^{2x})/(e^{2x)+14e^x+45} is 9ln|e^x+9|-5ln|e^x+5|+C

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integral of (4e^{2x})/(e^{2x)+14e^x+45}

\int \frac{4 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 45} d x

9 ln ∣ e^{x} + 9 ∣ - 5 ln ∣ e^{x} + 5 ∣ + C

Solution steps

\int \frac{4 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 45} 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} + 14 e ^{x} + 45} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 14 u + 45} d u

= 4 \cdot \int \frac{u}{u ^{2} + 14 u + 45} d u

Complete the square

u^{2} + 14 u + 45 : (u + 7)^{2} - 4

= 4 \cdot \int \frac{u}{( u + 7 ) ^{2} - 4} d u

Apply u-substitution

: \int \frac{v - 7}{v ^{2} - 4} d v

= 4 \cdot \int \frac{v - 7}{v ^{2} - 4} d v

Take the partial fraction of

\frac{v - 7}{v ^{2} - 4} : \frac{9}{4 ( v + 2 )} - \frac{5}{4 ( v - 2 )}

= 4 \cdot \int \frac{9}{4 ( v + 2 )} - \frac{5}{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{9}{4 ( v + 2 )} d v - \int \frac{5}{4 ( v - 2 )} d v)

\int \frac{9}{4 ( v + 2 )} d v = \frac{9}{4} ln ∣ v + 2 ∣

\int \frac{5}{4 ( v - 2 )} d v = \frac{5}{4} ln ∣ v - 2 ∣

= 4 (\frac{9}{4} ln ∣ v + 2 ∣ - \frac{5}{4} ln ∣ v - 2 ∣)

Substitute back

= 4 (\frac{9}{4} ln ∣ e^{x} + 7 + 2 ∣ - \frac{5}{4} ln ∣ e^{x} + 7 - 2 ∣)

Simplify

4 (\frac{9}{4} ln ∣ e^{x} + 7 + 2 ∣ - \frac{5}{4} ln ∣ e^{x} + 7 - 2 ∣) : 9 ln ∣ e^{x} + 9 ∣ - 5 ln ∣ e^{x} + 5 ∣

= 9 ln ∣ e^{x} + 9 ∣ - 5 ln ∣ e^{x} + 5 ∣

Add a constant to the solution

= 9 ln ∣ e^{x} + 9 ∣ - 5 ln ∣ e^{x} + 5 ∣ + C

\frac{d}{d x} ((\frac{5}{x} + x) (\frac{5}{x} - x))

y^{^{'}} = (x - 2) e^{- 2 y}

\int \frac{1}{9 e ^{- 6 x} + e ^{6 x}} d x

3 y^{^{''}} + 27 y = 0

\frac{32}{x}

What is the integral of (4e^{2x})/(e^{2x)+14e^x+45} ?
The integral of (4e^{2x})/(e^{2x)+14e^x+45} is 9ln|e^x+9|-5ln|e^x+5|+C