What is the integral of (x+9/2)e^{x^2+9x+1} ?

The integral of (x+9/2)e^{x^2+9x+1} is 1/2 e^{x^2+9x+1}+C

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integral of (x+9/2)e^{x^2+9x+1}

\int (x + \frac{9}{2}) e^{x^{2} + 9 x + 1} d x

\frac{1}{2} e^{x^{2} + 9 x + 1} + C

Solution steps

\int (x + \frac{9}{2}) e^{x^{2} + 9 x + 1} d x

Simplify

x + \frac{9}{2} : \frac{2 x + 9}{2}

= \int \frac{2 x + 9}{2} e^{x^{2} + 9 x + 1} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int (2 x + 9) e^{x^{2} + 9 x + 1} d x

Apply u-substitution

: \int e^{u} d u

= \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^{2} + 9 x + 1

= \frac{1}{2} e^{x^{2} + 9 x + 1}

Add a constant to the solution

= \frac{1}{2} e^{x^{2} + 9 x + 1} + C

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What is the integral of (x+9/2)e^{x^2+9x+1} ?
The integral of (x+9/2)e^{x^2+9x+1} is 1/2 e^{x^2+9x+1}+C