Can someone help me to evaluate this integral: Can someone help me with evaluating this integral:
$$\int_{1}^{2} \frac{2x^2-1} {\sqrt{x^2-1}}\, dx$$
I tried using integration by parts, integration by substitution....but nothing...
 A: Hint: 
Another useful subsitution that simplifies integrals with $\sqrt{x^2-1}$ is $x = \cosh u$. You then use the identities $\cosh^2 - 1 = \sinh^2$ and $\cosh' = \sinh$.
A: Hint: $$\int\frac{2x^2-1} {\sqrt{x^2-1}}\, dx = x\sqrt{x^2-1}+C$$ thus $$\int_{1}^{2} \frac{2x^2-1} {\sqrt{x^2-1}}\, dx=2\sqrt{4-1}-1\sqrt{1-1}\\=2\sqrt3$$
You can substitute $x^2-1=z$ to get $$\int\frac{2x^2-1} {\sqrt{x^2-1}}\, dx \\= \frac12\int\frac{2z+1} {\sqrt{z}}\, dz \\= \frac12\left(\int \sqrt{z}\, dz+\int \frac{1}{\sqrt{z}}\, dz\right)\\=\frac13\sqrt{z^3}+\sqrt{z}+C=\sqrt{z}\left(\frac{z}{3}+1\right)+C$$
So, making the final passage
$$\int_{1}^{2} \frac{2x^2-1} {\sqrt{x^2-1}}\, dx \\= \int_{0}^{3} \frac12\int\frac{2z+1} {\sqrt{z}}\, dz\\=\sqrt{3}\left(\frac{3}{3}+1\right)-\sqrt{0}\left(\frac{0}{3}+1\right)=2\sqrt3$$
A: HINT:
Start with Trigonometric Substitution  $$x=\sec\theta$$ 
to get $$I=\int_0^{\frac\pi3}(2\sec^2\theta-1)\frac{\sec\theta\tan\theta\ d\theta}{\tan\theta}=\int_0^{\frac\pi3}(2\sec^3\theta-\sec\theta)\ d\theta$$
Then use this
A: $$\frac{2x^2-1}{\sqrt{x^2-1}}=\frac{2(x^2-1)+1}{\sqrt{x^2-1}}=2\sqrt{x^2-1}+\frac1{\sqrt{x^2-1}}$$
Now use $\#1,\#8$  formulae of this
