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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...

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$\dfrac u{\sqrt{u}} = \sqrt{u}$ when $u\ne 0$. – Justin Apr 29 '14 at 18:04
up vote 2 down vote accepted

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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It's correct but how did you get it? – I Love Math Apr 29 '14 at 16:39
Well, once you get used to "see" derivative of the functions, sometimes it's easy to solve integrals..moreover I did it a similar one yesterday ;) – sirfoga Apr 29 '14 at 16:42
suppose that i can't find the primitive integral, so how should I go on? – I Love Math Apr 29 '14 at 16:44
You can substitute $x^2-1=z$ to get $$\int\frac{2x^2-1} {\sqrt{x^2-1}}\, dx \\= \int\frac{2z+1} {\sqrt{z}}\, dz \\= 2\int \sqrt{z}\, dz+\int \frac{1}{\sqrt{z}}\, dz\\=\frac43\sqrt{z^3}+2\sqrt{z}$$ – sirfoga Apr 29 '14 at 16:50
Well i get $6\sqrt{3}$... – I Love Math Apr 29 '14 at 16:58


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 $\cosh' = \sinh$.

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and also $2\cosh^2u-1=\cosh2u$. (+1) – Start wearing purple Apr 29 '14 at 16:45

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

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Now use $\#1,\#8$ formulae of this

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I can't use #8...notice that i have -1 under square root – I Love Math Apr 29 '14 at 16:49
@ILoveMath, Replace $a^2$ with $-b^2$ – lab bhattacharjee Apr 29 '14 at 16:52

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