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

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

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integral of e^{(x+1)^2}(x+1)

\int e^{(x + 1)^{2}} (x + 1) d x

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

Solution steps

\int e^{(x + 1)^{2}} (x + 1) d x

Apply u-substitution

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

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = (x + 1)^{2}

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

Add a constant to the solution

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

f (x) = \frac{1}{x} - 2, (- 1, - 3)

g (x) = x^{3} \cdot e^{x}

\int_{0}^{\frac{π}{6}} (3 sec^{2} (x)) d x

\int 3 cos (t) d t

x \to 3 π + lim (1 + csc (x))

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