Find the indefinite integral by using the substitution $$ x=4 \tan (\theta) $$. (Use $$ C $$ for th $$ \int \frac{4 x^{2}}{\left(16+x^{2}\right)^{2}} d x $$

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Find the integral $$ \int \frac{4x^{2}}{(16+x^{2})^{2}}dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int \frac{4x^{2}}{\left(16+x^{2}ight)^{2}} dx$$
- step1: Rewrite the expression:
  $$\int 4	imes \frac{x^{2}}{\left(16+x^{2}ight)^{2}} dx$$
- step2: Use properties of integrals:
  $$4	imes \int \frac{x^{2}}{\left(16+x^{2}ight)^{2}} dx$$
- step3: Rewrite the fraction:
  $$4	imes \int \left(-\frac{16}{\left(16+x^{2}ight)^{2}}+\frac{1}{16+x^{2}}ight) dx$$
- step4: Use properties of integrals:
  $$4\left(\int -\frac{16}{\left(16+x^{2}ight)^{2}} dx+\int \frac{1}{16+x^{2}} dxight)$$
- step5: Calculate:
  $$4	imes \int -\frac{16}{\left(16+x^{2}ight)^{2}} dx+4	imes \int \frac{1}{16+x^{2}} dx$$
- step6: Evaluate the integral:
  $$-\frac{2x}{16+x^{2}}-\frac{\arctan\left(\frac{x}{4}ight)}{2}+4	imes \int \frac{1}{16+x^{2}} dx$$
- step7: Evaluate the integral:
  $$-\frac{2x}{16+x^{2}}-\frac{\arctan\left(\frac{x}{4}ight)}{2}+\arctan\left(\frac{x}{4}ight)$$
- step8: Add the constant of integral C:
  $$-\frac{2x}{16+x^{2}}-\frac{\arctan\left(\frac{x}{4}ight)}{2}+\arctan\left(\frac{x}{4}ight) + C, C \in \mathbb{R}$$
The indefinite integral of $$ \frac{4x^{2}}{(16+x^{2})^{2}} $$ with respect to $$ x $$ is $$ -\frac{2x}{16+x^{2}}-\frac{\arctan(\frac{x}{4})}{2}+\arctan(\frac{x}{4}) + C $$, where $$ C $$ is an arbitrary constant.
The integral $$ \int \frac{4x^{2}}{(16+x^{2})^{2}}dx $$ equals $$ -\frac{2x}{16+x^{2}}-\frac{\arctan(\frac{x}{4})}{2}+\arctan(\frac{x}{4}) + C $$, where $$ C $$ is a constant.

Find the indefinite integral by using the substitution . (Use for th

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