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

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

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

\int e^{x - 1} d x

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

Solution steps

\int e^{x - 1} d x

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = x - 1

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

Add a constant to the solution

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

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(8.6) (8.1)

\int \frac{1}{25 - 81 x ^{2}} d x

e^{x}, - 1, 2

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

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