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

The integral of ((1-x))/(e^x)k is k(-(1-x)/(e^x)+1/(e^x))+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1 - ln ( u )}{u ^{2}} d u

= k \cdot \int \frac{1 - ln ( u )}{u ^{2}} d u

Apply Integration By Parts

: - \frac{1 - ln ( u )}{u} - \int \frac{1}{u ^{2}} d u

= k (- \frac{1 - ln ( u )}{u} - \int \frac{1}{u ^{2}} d u)

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

= k (- \frac{1 - ln ( u )}{u} - (- \frac{1}{u}))

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (+ (ln (x)))

y^{^{''}} - 8 y^{^{'}} - 15 y = 0

\frac{2 x}{5}

x \to 2 lim (3 x^{2} + 4 x - 3)

\int 2 (1 + 2 x)^{4} d x

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