Determine the following indefinite integral

Question

$\int \frac{dx}{x\sqrt{289x^2-289}}$

The only technique we have learned is the reverse power rule since we have just started this concept. How would I go about solving this with only the reverse power rule?

Answer 1

Using $17^2$=289 we can rewrite the integral as $\frac{1}{17}\int \frac{dx}{x\sqrt{x^2-1}}$ and then we use the substitution $y=\sqrt(x^2-1)$) to obtain $\int \frac{dy}{1+y^2}$ which is arctan(y).

Hence we have $\int \frac{dx}{x\sqrt{289x^2-289}}$=$\frac{arctan(sqrt{(x^2-1)})}{17} +C$. This is integration using inverse trigonometric functions.

Answer 2

This is actually an elementary antiderivative.

$\displaystyle \int\dfrac{dx}{x\sqrt{289x^2-289}} $

Bring out the constant $\sqrt{289}=17$

Now,

$ \displaystyle \dfrac{1}{17}\int\dfrac{dx}{x\sqrt{x^2-1}} $

Note that this is a standard antiderivative. $\dfrac{d}{dx}\sec^{-1} (x)=\dfrac{1}{x\sqrt{x^2-1}}$.

Here,$ \sec^{-1}(x)$ is the inverse of $\sec(x)$

So the integral is $\displaystyle \dfrac{1}{17} \sec^{-1} (x ) +C$