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

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

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integral of 4/(e^{1-2x)}

\int \frac{4}{e ^{1 - 2 x}} d x

2 e^{2 x - 1} + C

Solution steps

\int \frac{4}{e ^{1 - 2 x}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{e ^{1 - 2 x}} d x

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

= 4 \cdot \int \frac{1}{e ^{1}} \cdot \frac{1}{e ^{- 2 x}} d x

Take the constant out:

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

= 4 \cdot \frac{1}{e ^{1}} \cdot \int \frac{1}{e ^{- 2 x}} d x

Simplify

= 4 \cdot \frac{1}{e} \cdot \int \frac{1}{e ^{- 2 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

= 4 \cdot \frac{1}{e} \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x

= 4 \cdot \frac{1}{e} \cdot \frac{1}{2} e^{2 x}

Simplify

4 \cdot \frac{1}{e} \cdot \frac{1}{2} e^{2 x} : 2 e^{2 x - 1}

= 2 e^{2 x - 1}

Add a constant to the solution

= 2 e^{2 x - 1} + C

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