Integral of $$$y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right)$$$ with respect to $$$x$$$

The calculator will find the integral/antiderivative of $$$y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right)$$$ with respect to $$$x$$$, with steps shown.

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Find $$$\int y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right)\, dx$$$.

Solution

Simplify the integrand:

$${\color{red}{\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x}}} = {\color{red}{\int{2 x y^{\frac{9}{2}} \left(2 x^{2} - y\right) d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2 y^{\frac{9}{2}}$$$ and $$$f{\left(x \right)} = x \left(2 x^{2} - y\right)$$$:

$${\color{red}{\int{2 x y^{\frac{9}{2}} \left(2 x^{2} - y\right) d x}}} = {\color{red}{\left(2 y^{\frac{9}{2}} \int{x \left(2 x^{2} - y\right) d x}\right)}}$$

Let $$$u=2 x^{2} - y$$$.

Then $$$du=\left(2 x^{2} - y\right)^{\prime }dx = 4 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{4}$$$.

Thus,

$$2 y^{\frac{9}{2}} {\color{red}{\int{x \left(2 x^{2} - y\right) d x}}} = 2 y^{\frac{9}{2}} {\color{red}{\int{\frac{u}{4} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = u$$$:

$$2 y^{\frac{9}{2}} {\color{red}{\int{\frac{u}{4} d u}}} = 2 y^{\frac{9}{2}} {\color{red}{\left(\frac{\int{u d u}}{4}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\frac{y^{\frac{9}{2}} {\color{red}{\int{u d u}}}}{2}=\frac{y^{\frac{9}{2}} {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=\frac{y^{\frac{9}{2}} {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$

Recall that $$$u=2 x^{2} - y$$$:

$$\frac{y^{\frac{9}{2}} {\color{red}{u}}^{2}}{4} = \frac{y^{\frac{9}{2}} {\color{red}{\left(2 x^{2} - y\right)}}^{2}}{4}$$

Therefore,

$$\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x} = \frac{y^{\frac{9}{2}} \left(2 x^{2} - y\right)^{2}}{4}$$

Simplify:

$$\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x} = \frac{y^{\frac{9}{2}} \left(- 2 x^{2} + y\right)^{2}}{4}$$

Add the constant of integration:

$$\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x} = \frac{y^{\frac{9}{2}} \left(- 2 x^{2} + y\right)^{2}}{4}+C$$

Answer

$$$\int y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right)\, dx = \frac{y^{\frac{9}{2}} \left(- 2 x^{2} + y\right)^{2}}{4} + C$$$A


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