Integral of $$$i a g h o r^{3} t u w x^{8} - i f n t y$$$ with respect to $$$x$$$

The calculator will find the integral/antiderivative of $$$i a g h o r^{3} t u w x^{8} - i f n t y$$$ with respect to $$$x$$$, with steps shown.

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Find $$$\int \left(i a g h o r^{3} t u w x^{8} - i f n t y\right)\, dx$$$.

Solution

Integrate term by term:

$${\color{red}{\int{\left(i a g h o r^{3} t u w x^{8} - i f n t y\right)d x}}} = {\color{red}{\left(- \int{i f n t y d x} + \int{i a g h o r^{3} t u w x^{8} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=i f n t y$$$:

$$\int{i a g h o r^{3} t u w x^{8} d x} - {\color{red}{\int{i f n t y d x}}} = \int{i a g h o r^{3} t u w x^{8} d x} - {\color{red}{i f n t x y}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=i a g h o r^{3} t u w$$$ and $$$f{\left(x \right)} = x^{8}$$$:

$$- i f n t x y + {\color{red}{\int{i a g h o r^{3} t u w x^{8} d x}}} = - i f n t x y + {\color{red}{i a g h o r^{3} t u w \int{x^{8} d x}}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=8$$$:

$$i a g h o r^{3} t u w {\color{red}{\int{x^{8} d x}}} - i f n t x y=i a g h o r^{3} t u w {\color{red}{\frac{x^{1 + 8}}{1 + 8}}} - i f n t x y=i a g h o r^{3} t u w {\color{red}{\left(\frac{x^{9}}{9}\right)}} - i f n t x y$$

Therefore,

$$\int{\left(i a g h o r^{3} t u w x^{8} - i f n t y\right)d x} = \frac{i a g h o r^{3} t u w x^{9}}{9} - i f n t x y$$

Simplify:

$$\int{\left(i a g h o r^{3} t u w x^{8} - i f n t y\right)d x} = \frac{i t x \left(a g h o r^{3} u w x^{8} - 9 f n y\right)}{9}$$

Add the constant of integration:

$$\int{\left(i a g h o r^{3} t u w x^{8} - i f n t y\right)d x} = \frac{i t x \left(a g h o r^{3} u w x^{8} - 9 f n y\right)}{9}+C$$

Answer

$$$\int \left(i a g h o r^{3} t u w x^{8} - i f n t y\right)\, dx = \frac{i t x \left(a g h o r^{3} u w x^{8} - 9 f n y\right)}{9} + C$$$A