Mathematics

Evaluate the given integral.
$$\displaystyle \int { \cfrac { x+2 }{ \sqrt { { x }^{ 2 }-1 }  }  } dx$$


SOLUTION
$$I=\displaystyle\int{\dfrac{x+2}{\sqrt{{x}^{2}-1}}dx}$$

$$I=\displaystyle\int{\dfrac{x\,dx}{\sqrt{{x}^{2}-1}}}+2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$$

$$I=\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}+2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$$

Consider $$\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}$$

Let $$t={x}^{2}-1\Rightarrow\,dt=2x\,dx$$

$$\Rightarrow\,\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{\sqrt{t}}}$$

$$=\displaystyle\int{{t}^{-\frac{1}{2}}dt}$$

$$=\dfrac{{t}^{-\frac{1}{2}+1}}{-\dfrac{1}{2}+1}+c$$

$$=\dfrac{{t}^{\frac{1}{2}}}{\dfrac{1}{2}}+c$$

$$=2\sqrt{t}+c$$

$$=2\sqrt{{x}^{2}-1}+c$$ where $$t={x}^{2}-1$$

Consider $$2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$$

We know that $$\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-{a}^{2}}}}=\log{\left|x+\sqrt{{x}^{2}-{a}^{2}}\right|}+c$$

$$2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}=2\log{\left|x+\sqrt{{x}^{2}-1}\right|}+c$$

$$\therefore\,I=\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}+2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$$

       $$=2\sqrt{{x}^{2}-1}+2\log{\left|x+\sqrt{{x}^{2}-1}\right|}+c$$
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