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

\frac{d}{d x} (- \frac{11}{x ^{4}})

\frac{\partial}{\partial x} (e^{x} \cdot cos (y))

\int \frac{x}{x ^{2} + 3 x - 4} d x

\int \frac{x ^{3}}{5 x ^{2} + 1} d x

\frac{d}{d x} ((arctan (x))^{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