What is the integral of e^{-x^2-2x+7}(x+1) ?

The integral of e^{-x^2-2x+7}(x+1) is -1/2 e^{-x^2-2x+7}+C

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integral of e^{-x^2-2x+7}(x+1)

\int e^{- x^{2} - 2 x + 7} (x + 1) d x

- \frac{1}{2} e^{- x^{2} - 2 x + 7} + C

Solution steps

\int e^{- x^{2} - 2 x + 7} (x + 1) d x

Apply u-substitution

: \int e^{- u^{2} + 8} u d u

= \int e^{- u^{2} + 8} u d u

Apply u-substitution

: \int - \frac{e ^{v}}{2} d v

= \int - \frac{e ^{v}}{2} d v

Take the constant out:

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

= - \frac{1}{2} \cdot \int e^{v} d v

Use the common integral:

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

= - \frac{1}{2} e^{v}

Substitute back

= - \frac{1}{2} e^{- (x + 1)^{2} + 8}

Simplify

- \frac{1}{2} e^{- (x + 1)^{2} + 8} : - \frac{1}{2} e^{- x^{2} - 2 x + 7}

= - \frac{1}{2} e^{- x^{2} - 2 x + 7}

Add a constant to the solution

= - \frac{1}{2} e^{- x^{2} - 2 x + 7} + C

y^{^{''}} + 16 y = 3 sin (4 t), y (0) = 0, y^{^{'}} (0) = 0

y = x^{3} - 9 x, (- 3, 0)

\int_{0}^{1} 7 e^{x^{2}} d x

\frac{d}{d x} (\frac{2 x ^{2} - 1}{x 1 + x ^{2}})

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

What is the integral of e^{-x^2-2x+7}(x+1) ?
The integral of e^{-x^2-2x+7}(x+1) is -1/2 e^{-x^2-2x+7}+C