Integral of $$$\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1$$$

Find $$$\int \left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$\int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{x}}$$

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

$$x + {\color{red}{\int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x}}} = x + {\color{red}{\int{\left(x^{2} - 2 + \frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}}\right)d x}}}$$

Integrate term by term:

$$x + {\color{red}{\int{\left(x^{2} - 2 + \frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}}\right)d x}}} = x + {\color{red}{\left(- \int{2 d x} + \int{x^{2} d x} + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=2$$$:

$$x + \int{x^{2} d x} + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} - {\color{red}{\int{2 d x}}} = x + \int{x^{2} d x} + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} - {\color{red}{\left(2 x\right)}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$- x + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\int{x^{2} d x}}}=- x + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- x + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Perform partial fraction decomposition (steps can be seen »):

$$\frac{x^{3}}{3} - x + {\color{red}{\int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x}}} = \frac{x^{3}}{3} - x + {\color{red}{\int{\left(\frac{3}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}}}$$

Integrate term by term:

$$\frac{x^{3}}{3} - x + {\color{red}{\int{\left(\frac{3}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}}} = \frac{x^{3}}{3} - x + {\color{red}{\left(- \int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x} + \int{\frac{3}{x^{2} + 1} d x}\right)}}$$

To calculate the integral $$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}$$$, apply integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$ to the integral $$$\int{\frac{1}{x^{2} + 1} d x}$$$.

Let $$$\operatorname{u}=\frac{1}{x^{2} + 1}$$$ and $$$\operatorname{dv}=dx$$$.

Then $$$\operatorname{du}=\left(\frac{1}{x^{2} + 1}\right)^{\prime }dx=- \frac{2 x}{\left(x^{2} + 1\right)^{2}} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d x}=x$$$ (steps can be seen »).

Thus,

$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}=\frac{1}{x^{2} + 1} \cdot x-\int{x \cdot \left(- \frac{2 x}{\left(x^{2} + 1\right)^{2}}\right) d x}=\frac{x}{x^{2} + 1} - \int{\left(- \frac{2 x^{2}}{\left(x^{2} + 1\right)^{2}}\right)d x}$$

Strip out the constant:

$$\frac{x}{x^{2} + 1} - \int{\left(- \frac{2 x^{2}}{\left(x^{2} + 1\right)^{2}}\right)d x}=\frac{x}{x^{2} + 1} + 2 \int{\frac{x^{2}}{\left(x^{2} + 1\right)^{2}} d x}$$

Rewrite the numerator of the integrand as $$$x^{2}=x^{2}{\color{red}{+1}}{\color{red}{-1}}$$$ and split:

$$\frac{x}{x^{2} + 1} + 2 \int{\frac{x^{2}}{\left(x^{2} + 1\right)^{2}} d x}=\frac{x}{x^{2} + 1} + 2 \int{\left(- \frac{1}{\left(x^{2} + 1\right)^{2}} + \frac{x^{2} + 1}{\left(x^{2} + 1\right)^{2}}\right)d x}=\frac{x}{x^{2} + 1} + 2 \int{\left(\frac{1}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}$$

Split the integrals:

$$\frac{x}{x^{2} + 1} + 2 \int{\left(\frac{1}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}=\frac{x}{x^{2} + 1} - 2 \int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x} + 2 \int{\frac{1}{x^{2} + 1} d x}$$

Thus, we get following simple linear equation with respect to the integral:

$$\int{\frac{1}{x^{2} + 1} d x}=\frac{x}{x^{2} + 1} + 2 \int{\frac{1}{x^{2} + 1} d x} - 2 {\color{red}{\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}}}$$

Solving it we obtain that

$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}=\frac{x}{2 \left(x^{2} + 1\right)} + \frac{\int{\frac{1}{x^{2} + 1} d x}}{2}$$

Therefore,

$$\frac{x^{3}}{3} - x + \int{\frac{3}{x^{2} + 1} d x} - {\color{red}{\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}}} = \frac{x^{3}}{3} - x + \int{\frac{3}{x^{2} + 1} d x} - {\color{red}{\left(\frac{x}{2 \left(x^{2} + 1\right)} + \frac{\int{\frac{1}{x^{2} + 1} d x}}{2}\right)}}$$

The integral of $$$\frac{1}{x^{2} + 1}$$$ is $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

$$\frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} + \int{\frac{3}{x^{2} + 1} d x} - \frac{{\color{red}{\int{\frac{1}{x^{2} + 1} d x}}}}{2} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} + \int{\frac{3}{x^{2} + 1} d x} - \frac{{\color{red}{\operatorname{atan}{\left(x \right)}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2} + 1}$$$:

$$\frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + {\color{red}{\int{\frac{3}{x^{2} + 1} d x}}} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + {\color{red}{\left(3 \int{\frac{1}{x^{2} + 1} d x}\right)}}$$

The integral of $$$\frac{1}{x^{2} + 1}$$$ is $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

$$\frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + 3 {\color{red}{\int{\frac{1}{x^{2} + 1} d x}}} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + 3 {\color{red}{\operatorname{atan}{\left(x \right)}}}$$

Therefore,

$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} + \frac{5 \operatorname{atan}{\left(x \right)}}{2}$$

Simplify:

$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)}$$

Add the constant of integration:

$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)}+C$$

$$$\int \left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)\, dx = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)} + C$$$A