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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{v} \cdot 2 v d v

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

Take the constant out:

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

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

Apply Integration By Parts

: e^{v} v - \int e^{v} d v

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

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

= - \frac{1}{2} \cdot 2 (e^{v} v - e^{v})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int e^{x} 17 + e^{x} d x

n = 1 \sum \infty tan (n)

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

\int \frac{x - 6}{x ^{2} - 6 x + 8} d x

\frac{d}{d x} (cos (1))

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