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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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