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

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

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

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

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

Solution steps

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

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