Resuelva la siguiente integral $$ \int_{0}^{3} \frac{8 x}{\sqrt{x^{2}+16}} d x=\square $$

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Calculate the integral $$ \int_{0}^{3} \frac{8x}{\sqrt{x^{2}+16}} dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{0}^{3} \frac{8x}{\sqrt{x^{2}+16}} dx$$
- step1: Evaluate the integral:
  $$\int \frac{8x}{\sqrt{x^{2}+16}} dx$$
- step2: Rewrite the expression:
  $$\int 8	imes \frac{x}{\sqrt{x^{2}+16}} dx$$
- step3: Use properties of integrals:
  $$8	imes \int \frac{x}{\sqrt{x^{2}+16}} dx$$
- step4: Rewrite the expression:
  $$8	imes \int \frac{x}{\left(x^{2}+16ight)^{\frac{1}{2}}} dx$$
- step5: Use the substitution $$dx=\frac{1}{2x} dt$$ to transform the integral$$:$$
  $$8	imes \int \frac{x}{\left(x^{2}+16ight)^{\frac{1}{2}}}	imes \frac{1}{2x} dt$$
- step6: Simplify:
  $$8	imes \int \frac{1}{2\left(x^{2}+16ight)^{\frac{1}{2}}} dt$$
- step7: Use the substitution $$t=x^{2}$$ to transform the integral$$:$$
  $$8	imes \int \frac{1}{2\left(t+16ight)^{\frac{1}{2}}} dt$$
- step8: Rewrite the expression:
  $$8	imes \int \frac{1}{2}	imes \frac{1}{\left(t+16ight)^{\frac{1}{2}}} dt$$
- step9: Use properties of integrals:
  $$8	imes \frac{1}{2}	imes \int \frac{1}{\left(t+16ight)^{\frac{1}{2}}} dt$$
- step10: Multiply the numbers:
  $$4	imes \int \frac{1}{\left(t+16ight)^{\frac{1}{2}}} dt$$
- step11: Use the substitution $$dt=1 dv$$ to transform the integral$$:$$
  $$4	imes \int \frac{1}{\left(t+16ight)^{\frac{1}{2}}}	imes 1 dv$$
- step12: Simplify:
  $$4	imes \int \frac{1}{\left(t+16ight)^{\frac{1}{2}}} dv$$
- step13: Use the substitution $$v=t+16$$ to transform the integral$$:$$
  $$4	imes \int \frac{1}{v^{\frac{1}{2}}} dv$$
- step14: Evaluate the integral:
  $$4	imes \frac{v^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$$
- step15: Simplify:
  $$4	imes \frac{v^{\frac{1}{2}}}{\frac{1}{2}}$$
- step16: Multiply the terms:
  $$\frac{4v^{\frac{1}{2}}}{\frac{1}{2}}$$
- step17: Simplify:
  $$8v^{\frac{1}{2}}$$
- step18: Substitute back:
  $$8\left(t+16ight)^{\frac{1}{2}}$$
- step19: Substitute back:
  $$8\left(x^{2}+16ight)^{\frac{1}{2}}$$
- step20: Return the limits:
  $$\left(8\left(x^{2}+16ight)^{\frac{1}{2}}ight)\bigg |_{0}^{3}$$
- step21: Calculate the value:
  $$8$$
La integral $$ \int_{0}^{3} \frac{8 x}{\sqrt{x^{2}+16}} d x $$ se evalúa en 8.
La integral $$ \int_{0}^{3} \frac{8x}{\sqrt{x^{2}+16}} dx $$ es igual a 8.

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