Integral; How is this step done?

Question

I've been attempting several other integrals and have a hit a massive wall in terms of this question:

$$\int\frac{x^2}{x^2-4}dx$$

I tried looking into substitutions to perform but couldn't find any useful ones on my own, so after consulting an online calculator, the following method was used:

How is this performed? I cannot understand how "adding 0" in the numerator yields the fraction $\int(\frac{4}{x^2-4}+1)dx$. Can anyone help shed light on this? I feel like it's very basic, but I haven't a clue how it was done :S

Answer 1

Alternatively, consider the general way of adding fractions together $$\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{db}\, \qquad (b, d) \neq 0$$

The answer given considers \begin{align} 1+\frac{4}{x^{2}-4} &= \frac{1}{1}+\frac{4}{x^{2}-4} \\ &= \frac{(1)(x^{2}-4)+4(1)}{(1)(x^{2}-4)} \\ &= \frac{x^{2}-4+4}{x^{2}-4} \\ &= \frac{x^{2}}{x^{2}-4}\, \qquad (\text{QED}) \end{align}

Answer 2

Follow steps $$\int \frac { x^{ 2 } }{ x^{ 2 }-4 } dx=\int \frac { x^{ 2 }-4+4 }{ x^{ 2 }-4 } dx=\int \left( \frac { x^{ 2 }-4 }{ x^{ 2 }-4 } +\frac { 4 }{ x^{ 2 }-4 } \right) dx=\\ =\int { \left( 1+\frac { 4 }{ x^{ 2 }-4 } \right) dx= } \int { dx } +4\int { \frac { dx }{ { x }^{ 2 }-4 } } $$

Answer 3

HINT: $$\frac{x^2}{x^2-4}=\frac{x^2-4+4}{x^2-4}=1+\frac{4}{x^2-4}=1+\frac{4}{(x-2)(x+2)}$$