What is the integral of ((e^x-2)e^x)/(e^x+1) ?

The integral of ((e^x-2)e^x)/(e^x+1) is -3ln|e^x+1|+e^x+C

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integral of ((e^x-2)e^x)/(e^x+1)

\int \frac{( e ^{x} - 2 ) e ^{x}}{e ^{x} + 1} d x

- 3 ln ∣ e^{x} + 1 ∣ + e^{x} + C

Solution steps

\int \frac{( e ^{x} - 2 ) e ^{x}}{e ^{x} + 1} d x

Apply u-substitution

: \int \frac{u - 2}{u + 1} d u

= \int \frac{u - 2}{u + 1} d u

\frac{u - 2}{u + 1} = - \frac{3}{u + 1} + 1

= \int - \frac{3}{u + 1} + 1 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

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

\int \frac{3}{u + 1} d u = 3 ln ∣ u + 1 ∣

\int 1 d u = u

= - 3 ln ∣ u + 1 ∣ + u

Substitute back

u = e^{x}

= - 3 ln ∣ e^{x} + 1 ∣ + e^{x}

Add a constant to the solution

= - 3 ln ∣ e^{x} + 1 ∣ + e^{x} + C

\frac{d ^{2}}{d x ^{2}} (x)

x, - x, 1, 4

sin (2 x)

\int x + \frac{x ^{2}}{2} + \frac{5 x ^{3}}{3} d x

\int_{0}^{\infty} \frac{5}{x ^{2} + 1} d x

What is the integral of ((e^x-2)e^x)/(e^x+1) ?
The integral of ((e^x-2)e^x)/(e^x+1) is -3ln|e^x+1|+e^x+C