What is the integral of 1/(4e^x) ?

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

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integral of 1/(4e^x)

\int \frac{1}{4 e ^{x}} d x

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

Solution steps

\int \frac{1}{4 e ^{x}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{e ^{x}} d x

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{4} (- \frac{1}{u})

Substitute back

u = e^{x}

= \frac{1}{4} (- \frac{1}{e ^{x}})

Simplify

\frac{1}{4} (- \frac{1}{e ^{x}}) : - \frac{1}{4 e ^{x}}

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

Add a constant to the solution

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

\int 19 sec^{- 2} (x t a) n^{3} x d x

y^{^{'}} + 6 y = 1, y (0) = 1

x^{2} + x y - y^{2} = 4

16 y^{^{''}} + 72 y^{^{'}} + 87 y = 0, y (0) = 2, y^{^{'}} (0) = 4

\frac{d v}{d x} - 8 v = 10