What is the integral of 2xyze^{x^2} ?

The integral of 2xyze^{x^2} is yze^{x^2}+C

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integral of 2xyze^{x^2}

\int 2 x yz e^{x^{2}} d x

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

Solution steps

\int 2 x yz e^{x^{2}} d x

Take the constant out:

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

= 2 yz \cdot \int e^{x^{2}} x d x

Apply u-substitution

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

= 2 yz \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

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

Use the common integral:

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

= 2 yz \frac{1}{2} e^{u}

Substitute back

u = x^{2}

= 2 yz \frac{1}{2} e^{x^{2}}

Simplify

2 yz \frac{1}{2} e^{x^{2}} : yz e^{x^{2}}

= yz e^{x^{2}}

Add a constant to the solution

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

\int \frac{4}{1 + t ^{2}} d t

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

\int (y) d y

f (x) = e^{\frac{x}{3}}

y^{^{''}} - 8 y^{^{'}} + 25 y = 0

What is the integral of 2xyze^{x^2} ?
The integral of 2xyze^{x^2} is yze^{x^2}+C