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$$\int{\frac{1}{x\sqrt{25x^2-1}}}\,dx$$

Let $x=\frac{1}{5}u$

Now when I substite it and simplify I get

$$\int{\frac{1}{u\sqrt{u^2-1}}}\, du$$

There is a trig identity which says that this is equal to $$\sec^{-1} u$$

But this is incorrect. I can't figure this out. Where did I go wrong?

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3 Answers

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$$\int\frac{dx}{x\sqrt{25x^2-1}}=\int\frac{25x}{25x^2\sqrt{25x^2-1}}dx$$

Let $\sqrt{25x^2-1}=u\implies\dfrac{50x\ dx}{2\sqrt{25x^2-1}}=du$ and $25x^2-1=u^2\iff25x^2=u^2+1$

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It should be: $\displaystyle \int \dfrac{du}{u\sqrt{u^2-1}} $, which equals to $\sec^{-1} u + C = \sec^{-1} (5x) + C$

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If $u=\sec\theta$ the $\sqrt{u^2-1}=\tan\theta$ and $\dfrac{du}u = \tan\theta\,d\theta$, so the integral is $$\int 1\,d\theta = \theta+C = \sec^{-1}u+C.$$

This works for $\theta$ in the first quadrant. In other quadrants, you need to think about "$\pm$' issues.

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