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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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

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

Apply u-substitution

= \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

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

Simplify

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

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

Add a constant to the solution

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

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