What is the integral of (5e^x)/(e^{2x)+2e^x+1} ?

The integral of (5e^x)/(e^{2x)+2e^x+1} is -5/(e^x+1)+C

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integral of (5e^x)/(e^{2x)+2e^x+1}

\int \frac{5 e ^{x}}{e ^{2 x} + 2 e ^{x} + 1} d x

- \frac{5}{e ^{x} + 1} + C

Solution steps

\int \frac{5 e ^{x}}{e ^{2 x} + 2 e ^{x} + 1} 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 ^{x}}{e ^{2 x} + 2 e ^{x} + 1} d x

Factor

\frac{e ^{x}}{e ^{2 x} + 2 e ^{x} + 1}

= 5 \cdot \int \frac{e ^{x}}{( e ^{x} + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= 5 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 5 (- \frac{1}{u})

Substitute back

u = e^{x} + 1

= 5 (- \frac{1}{e ^{x} + 1})

Simplify

5 (- \frac{1}{e ^{x} + 1}) : - \frac{5}{e ^{x} + 1}

= - \frac{5}{e ^{x} + 1}

Add a constant to the solution

= - \frac{5}{e ^{x} + 1} + C

(- 4, 0), (1, - 5)

\frac{d}{d x} (x^{2} + 2 y^{2} - 4)

(y - x) d x + 4 x d y = 0

\int \frac{1}{2 x - 1} d x

ln ∣ sin (x) ∣

What is the integral of (5e^x)/(e^{2x)+2e^x+1} ?
The integral of (5e^x)/(e^{2x)+2e^x+1} is -5/(e^x+1)+C