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

The integral of (2-e^x)^2 is 4x-4e^x+1/2 e^{2x}+C

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integral of (2-e^x)^2

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

4 x - 4 e^{x} + \frac{1}{2} e^{2 x} + C

Solution steps

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

Apply u-substitution

: \int \frac{( 2 - u ) ^{2}}{u} d u

= \int \frac{( 2 - u ) ^{2}}{u} d u

Expand

\frac{( 2 - u ) ^{2}}{u} : \frac{4}{u} - 4 + u

= \int \frac{4}{u} - 4 + u 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{4}{u} d u - \int 4 d u + \int u d u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

\int 4 d u = 4 u

\int u d u = \frac{u ^{2}}{2}

= 4 ln ∣ u ∣ - 4 u + \frac{u ^{2}}{2}

Substitute back

u = e^{x}

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

Simplify

4 ln ∣ e^{x} ∣ - 4 e^{x} + \frac{( e ^{x} ) ^{2}}{2} : 4 x - 4 e^{x} + \frac{1}{2} e^{2 x}

= 4 x - 4 e^{x} + \frac{1}{2} e^{2 x}

Add a constant to the solution

= 4 x - 4 e^{x} + \frac{1}{2} e^{2 x} + C

f (x) = x^{2} (1 - 5 x)

\int \frac{30}{( 225 x ^{2} + 1 ) ^{2}} d x

\frac{10}{( 1 + e ^{- x} )}

\frac{\partial}{\partial x} (e^{\frac{2}{x}})

\int_{3}^{\infty} 1

What is the integral of (2-e^x)^2 ?
The integral of (2-e^x)^2 is 4x-4e^x+1/2 e^{2x}+C