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

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

integral of e^{2sqrt(x)}

\int e^{2 x} d x

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

Solution steps

\int e^{2 x} d x

Apply u-substitution

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

Apply Integration By Parts

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

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

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{z ^{2} + 4} d z

\frac{d}{d x} (\frac{sin ( x )}{1 + 2 cos ( x )})

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

\int \frac{z ^{2}}{z ^{3} + 1} d z

d er i v a t i v e - 9 (5 x^{2} + 4)^{- 6}

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