What is the integral of (4x^3-6x)/(e^{2x^2)} ?

The integral of (4x^3-6x)/(e^{2x^2)} is (-x^2+1)/(e^{2x^2)}+C

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integral of (4x^3-6x)/(e^{2x^2)}

\int \frac{4 x ^{3} - 6 x}{e ^{2 x^{2}}} d x

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

Solution steps

\int \frac{4 x ^{3} - 6 x}{e ^{2 x^{2}}} d x

Apply u-substitution

: \int \frac{u - 3}{2 e ^{u}} d u

= \int \frac{u - 3}{2 e ^{u}} 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 \frac{u - 3}{e ^{u}} d u

Expand

\frac{u - 3}{e ^{u}} : \frac{u}{e ^{u}} - \frac{3}{e ^{u}}

= \frac{1}{2} \cdot \int \frac{u}{e ^{u}} - \frac{3}{e ^{u}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{u}{e ^{u}} d u - \int \frac{3}{e ^{u}} d u)

\int \frac{u}{e ^{u}} d u = - \frac{u}{e ^{u}} - \frac{1}{e ^{u}}

\int \frac{3}{e ^{u}} d u = - \frac{3}{e ^{u}}

= \frac{1}{2} (- \frac{u}{e ^{u}} - \frac{1}{e ^{u}} - (- \frac{3}{e ^{u}}))

Substitute back

u = 2 x^{2}

= \frac{1}{2} (- \frac{2 x ^{2}}{e ^{2 x^{2}}} - \frac{1}{e ^{2 x^{2}}} - (- \frac{3}{e ^{2 x^{2}}}))

Simplify

\frac{1}{2} (- \frac{2 x ^{2}}{e ^{2 x^{2}}} - \frac{1}{e ^{2 x^{2}}} - (- \frac{3}{e ^{2 x^{2}}})) : \frac{- x ^{2} + 1}{e ^{2 x^{2}}}

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x^{2} + 1)

x \to 1 lim (\frac{1}{( x - 1 ) ^{4}})

\int_{- 2}^{0} \frac{x}{( x ^{2} + 3 ) ^{2}} d x

(\frac{1}{4 x})^{^{'}}

\frac{d}{d x} (4 x + 8 x^{\frac{3}{4}})

What is the integral of (4x^3-6x)/(e^{2x^2)} ?
The integral of (4x^3-6x)/(e^{2x^2)} is (-x^2+1)/(e^{2x^2)}+C