Integral of $$$\frac{4 t^{2} x^{2}}{\left(t^{2} + x^{2}\right)^{2}}$$$ with respect to $$$x$$$

Find $$$\int \frac{4 t^{2} x^{2}}{\left(t^{2} + x^{2}\right)^{2}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4 t^{2}$$$ and $$$f{\left(x \right)} = \frac{x^{2}}{\left(t^{2} + x^{2}\right)^{2}}$$$:

$${\color{red}{\int{\frac{4 t^{2} x^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x}}} = {\color{red}{\left(4 t^{2} \int{\frac{x^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x}\right)}}$$

Perform partial fraction decomposition:

$$4 t^{2} {\color{red}{\int{\frac{x^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x}}} = 4 t^{2} {\color{red}{\int{\left(- \frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} + \frac{1}{t^{2} + x^{2}}\right)d x}}}$$

Integrate term by term:

$$4 t^{2} {\color{red}{\int{\left(- \frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} + \frac{1}{t^{2} + x^{2}}\right)d x}}} = 4 t^{2} {\color{red}{\left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + \int{\frac{1}{t^{2} + x^{2}} d x}\right)}}$$

Let $$$u=\frac{x}{\left|{t}\right|}$$$.

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

The integral can be rewritten as

$$4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + {\color{red}{\int{\frac{1}{t^{2} + x^{2}} d x}}}\right) = 4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + {\color{red}{\int{\frac{\left|{t}\right|}{t^{2} \left(u^{2} + 1\right)} d u}}}\right)$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{\left|{t}\right|}{t^{2}}$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$:

$$4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + {\color{red}{\int{\frac{\left|{t}\right|}{t^{2} \left(u^{2} + 1\right)} d u}}}\right) = 4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + {\color{red}{\frac{\left|{t}\right| \int{\frac{1}{u^{2} + 1} d u}}{t^{2}}}}\right)$$

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

$$4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + \frac{\left|{t}\right| {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{t^{2}}\right) = 4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + \frac{\left|{t}\right| {\color{red}{\operatorname{atan}{\left(u \right)}}}}{t^{2}}\right)$$

Recall that $$$u=\frac{x}{\left|{t}\right|}$$$:

$$4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + \frac{\left|{t}\right| \operatorname{atan}{\left({\color{red}{u}} \right)}}{t^{2}}\right) = 4 t^{2} \left(- \int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} + \frac{\left|{t}\right| \operatorname{atan}{\left({\color{red}{\frac{x}{\left|{t}\right|}}} \right)}}{t^{2}}\right)$$

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

$$4 t^{2} \left(- {\color{red}{\int{\frac{t^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x}}} + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}\right) = 4 t^{2} \left(- {\color{red}{t^{2} \int{\frac{1}{\left(t^{2} + x^{2}\right)^{2}} d x}}} + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}\right)$$

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

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

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

The integral becomes

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

Strip out the constant:

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

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

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

Split the integrals:

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

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

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

Solving it we obtain that

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

Therefore,

$$4 t^{2} \left(- t^{2} {\color{red}{\int{\frac{1}{\left(t^{2} + x^{2}\right)^{2}} d x}}} + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}\right) = 4 t^{2} \left(- t^{2} {\color{red}{\left(\frac{x}{2 t^{2} \left(t^{2} + x^{2}\right)} + \frac{\int{\frac{1}{t^{2} + x^{2}} d x}}{2 t^{2}}\right)}} + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}\right)$$

The integral $$$\int{\frac{1}{t^{2} + x^{2}} d x}$$$ was already calculated:

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

Therefore,

$$4 t^{2} \left(- t^{2} \left(\frac{x}{2 t^{2} \left(t^{2} + x^{2}\right)} + \frac{{\color{red}{\int{\frac{1}{t^{2} + x^{2}} d x}}}}{2 t^{2}}\right) + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}\right) = 4 t^{2} \left(- t^{2} \left(\frac{x}{2 t^{2} \left(t^{2} + x^{2}\right)} + \frac{{\color{red}{\frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}}}}{2 t^{2}}\right) + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}\right)$$

Therefore,

$$\int{\frac{4 t^{2} x^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} = 4 t^{2} \left(- t^{2} \left(\frac{x}{2 t^{2} \left(t^{2} + x^{2}\right)} + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{2 t^{4}}\right) + \frac{\left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}}{t^{2}}\right)$$

Simplify:

$$\int{\frac{4 t^{2} x^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} = \frac{2 \left(- t^{2} x + \left(t^{2} + x^{2}\right) \left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}\right)}{t^{2} + x^{2}}$$

Add the constant of integration:

$$\int{\frac{4 t^{2} x^{2}}{\left(t^{2} + x^{2}\right)^{2}} d x} = \frac{2 \left(- t^{2} x + \left(t^{2} + x^{2}\right) \left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}\right)}{t^{2} + x^{2}}+C$$

$$$\int \frac{4 t^{2} x^{2}}{\left(t^{2} + x^{2}\right)^{2}}\, dx = \frac{2 \left(- t^{2} x + \left(t^{2} + x^{2}\right) \left|{t}\right| \operatorname{atan}{\left(\frac{x}{\left|{t}\right|} \right)}\right)}{t^{2} + x^{2}} + C$$$A