<s>[INST] 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...
 [/INST] 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
$$\int_{1}^{2} \frac{2x^2-1} {\sqrt{x^2-1}}\, dx \\= \int_{0}^{3} \frac12\int\frac{2z+1} {\sqrt{z}}\, dz\\=\sqrt{3}\left(\frac{3}{3}+1\right)-\sqrt{0}\left(\frac{0}{3}+1\right)=2\sqrt3$$
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