f $$ \int_{-2}^{0} \frac{8}{(2-x)\left(4+x^{2}\right)} \mathrm{d} x $$

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Calculate the integral $$ \int_{-2}^{0} \frac{8}{(2-x)(4+x^{2})} \,dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{-2}^{0} \frac{8}{\left(2-xight)\left(4+x^{2}ight)} dx$$
- step1: Multiply the terms:
  $$\int_{-2}^{0} \frac{8}{8+2x^{2}-4x-x^{3}} dx$$
- step2: Evaluate the integral:
  $$\int \frac{8}{8+2x^{2}-4x-x^{3}} dx$$
- step3: Rewrite the expression:
  $$\int 8	imes \frac{1}{8+2x^{2}-4x-x^{3}} dx$$
- step4: Use properties of integrals:
  $$8	imes \int \frac{1}{8+2x^{2}-4x-x^{3}} dx$$
- step5: Rewrite the fraction:
  $$8	imes \int \left(\frac{1}{16-8x}+\frac{2+x}{32+8x^{2}}ight) dx$$
- step6: Use properties of integrals:
  $$8\left(\int \frac{1}{16-8x} dx+\int \frac{2+x}{32+8x^{2}} dxight)$$
- step7: Calculate:
  $$8	imes \int \frac{1}{16-8x} dx+8	imes \int \frac{2+x}{32+8x^{2}} dx$$
- step8: Evaluate the integral:
  $$-\ln{\left(\left|x-2ight|ight)}+8	imes \int \frac{2+x}{32+8x^{2}} dx$$
- step9: Evaluate the integral:
  $$-\ln{\left(\left|x-2ight|ight)}+\arctan\left(\frac{x}{2}ight)+\frac{1}{2}\ln{\left(x^{2}+4ight)}$$
- step10: Add the terms:
  $$\ln{\left(\frac{\left(x^{2}+4ight)^{\frac{1}{2}}}{\left|x-2ight|}ight)}+\arctan\left(\frac{x}{2}ight)$$
- step11: Return the limits:
  $$\left(\ln{\left(\frac{\left(x^{2}+4ight)^{\frac{1}{2}}}{\left|x-2ight|}ight)}+\arctan\left(\frac{x}{2}ight)ight)\bigg |_{-2}^{0}$$
- step12: Calculate the value:
  $$\frac{2\ln{\left(2ight)}+\pi }{4}$$
The integral of $$ \frac{8}{(2-x)(4+x^{2})} $$ from -2 to 0 is $$ \frac{2\ln{(2)}+\pi}{4} $$.
The integral evaluates to $$ \frac{2\ln{(2)}+\pi}{4} $$.

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