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

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

Popular >

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

\int e^{1 - x} d x

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

Solution steps

\int e^{1 - x} d x

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Apply u-substitution

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

= - \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

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

Apply Integration By Parts

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

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

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

= - 2 (e^{v} v - e^{v})

Substitute back

= - 2 (e^{1 - x} 1 - x - e^{1 - x})

Add a constant to the solution

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

\int_{0}^{3} \frac{1}{9 + x ^{2}} d x

\int_{- 2}^{0} x^{3} - 4 x d x

\frac{\partial}{\partial x} (60 e^{x^{6} y} x^{5} y)

\int_{4}^{5} \frac{x}{x - 4} d x

\frac{\partial}{\partial x} ((y^{2} - 2)^{3})

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