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

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

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

integral

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

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

Solution steps

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

Take the constant out:

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

= \frac{1}{e ^{2}} \cdot \int \frac{1 - x ^{2}}{x} d x

Expand

\frac{1 - x ^{2}}{x} : \frac{1}{x} - x

= \frac{1}{e ^{2}} \cdot \int \frac{1}{x} - x d x

Apply the Sum Rule:

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= \frac{1}{e ^{2}} (ln ∣ x ∣ - \frac{x ^{2}}{2})

Add a constant to the solution

= \frac{1}{e ^{2}} (ln ∣ x ∣ - \frac{x ^{2}}{2}) + C

\frac{x ^{- 2}}{3}

sin (2 x) + cos (x)

\frac{1}{( x ^{2} + y ^{2} + z ^{2} )}

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

(\frac{1}{x}) sin (\frac{1}{x})

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