What is the integral of bxe^{3xy} ?

The integral of bxe^{3xy} is (be^{3xy})/3+C

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integral of bxe^{3xy}

\int b x e^{3 x y} d y

\frac{b e ^{3 x y}}{3} + C

Solution steps

\int b x e^{3 x y} d y

Take the constant out:

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

= b x \cdot \int e^{3 x y} d y

Apply u-substitution

: \int \frac{e ^{u}}{3 x} d u

= b x \cdot \int \frac{e ^{u}}{3 x} d u

Take the constant out:

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

= b x \frac{1}{3 x} \cdot \int e^{u} d u

Use the common integral:

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

= b x \frac{1}{3 x} e^{u}

Substitute back

u = 3 x y

= b x \frac{1}{3 x} e^{3 x y}

Simplify

b x \frac{1}{3 x} e^{3 x y} : \frac{b e ^{3 x y}}{3}

= \frac{b e ^{3 x y}}{3}

Add a constant to the solution

= \frac{b e ^{3 x y}}{3} + C

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