Solve ∫ frac{(2x^{2}y^{4})(4x^2y^4)(3x)}{3x^-3y^2}=

$$\int \frac { ( 2 x ^ { 2 } y ^ { 4 } ) ( 4 x ^ { 2 } y ^ { 4 } ) ( 3 x ) } { 3 x ^ { - 3 } y ^ { 2 } } =$$

Evaluate

$$\frac{8y^{6}x^{9}}{9}+С$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$\int \frac{2x^{4}y^{4}\times 4y^{4}\times 3x}{3x^{-3}y^{2}}\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 4 and 1 to get 5.

$$\int \frac{2x^{5}y^{4}\times 4y^{4}\times 3}{3x^{-3}y^{2}}\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 4 and 4 to get 8.

$$\int \frac{2x^{5}y^{8}\times 4\times 3}{3x^{-3}y^{2}}\mathrm{d}x$$

Cancel out 3y^{2} in both numerator and denominator.

$$\int \frac{2\times 4x^{5}y^{6}}{x^{-3}}\mathrm{d}x$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent.

$$\int 2\times 4y^{6}x^{8}\mathrm{d}x$$

Multiply 2 and 4 to get 8.

$$\int 8y^{6}x^{8}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$8y^{6}\int x^{8}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{8}\mathrm{d}x with \frac{x^{9}}{9}.

$$8y^{6}\times \frac{x^{9}}{9}$$

Simplify.

$$\frac{8y^{6}x^{9}}{9}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{8y^{6}x^{9}}{9}+С$$

Differentiate w.r.t. x

$$8y^{6}x^{8}$$