What is the integral of 9xe^{5x^2} ?

The integral of 9xe^{5x^2} is 9/10 e^{5x^2}+C

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integral of 9xe^{5x^2}

\int 9 x e^{5 x^{2}} d x

\frac{9}{10} e^{5 x^{2}} + C

Solution steps

\int 9 x e^{5 x^{2}} d x

Take the constant out:

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

= 9 \cdot \int x e^{5 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{10} d u

= 9 \cdot \int \frac{e ^{u}}{10} d u

Take the constant out:

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

= 9 \cdot \frac{1}{10} \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{10} e^{u}

Substitute back

u = 5 x^{2}

= 9 \cdot \frac{1}{10} e^{5 x^{2}}

Simplify

9 \cdot \frac{1}{10} e^{5 x^{2}} : \frac{9}{10} e^{5 x^{2}}

= \frac{9}{10} e^{5 x^{2}}

Add a constant to the solution

= \frac{9}{10} e^{5 x^{2}} + C

\frac{d}{d x} (A (e^{x} + e^{x} x))

x^{2} + x y + 2 y^{2} = 4, (- 2, 1)

9 y^{^{''}} + 6 y^{^{'}} + y = 0, y (- 3) = - 2, y^{^{'}} (1) = - 1

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

y^{^{'}} = 1 + x^{2} + y^{2} + x^{2} y^{2}

What is the integral of 9xe^{5x^2} ?
The integral of 9xe^{5x^2} is 9/10 e^{5x^2}+C