What is the integral of-4xe^{x^2} ?

The integral of-4xe^{x^2} is -2e^{x^2}+C

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integral of-4xe^{x^2}

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

- 2 e^{x^{2}} + C

Solution steps

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

Take the constant out:

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

= - 4 \cdot \int x e^{x^{2}} d x

Apply u-substitution

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

= - 4 \cdot \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

= - 4 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= - 4 \cdot \frac{1}{2} e^{u}

Substitute back

u = x^{2}

= - 4 \cdot \frac{1}{2} e^{x^{2}}

Simplify

- 4 \cdot \frac{1}{2} e^{x^{2}} : - 2 e^{x^{2}}

= - 2 e^{x^{2}}

Add a constant to the solution

= - 2 e^{x^{2}} + C

\frac{3}{( s - 1 ) ^{2}}

\int \frac{48 x ^{2}}{( x - 15 ) ( x + 5 ) ^{2}} d x

x \to - 3 lim (4 x^{2} + 3 x + 2)

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

y = \frac{2 x + 1}{x}