Evaluate $\int \frac{\mathrm{d}x}{x\sqrt{49x^2-1}}$

I keep getting the wrong answer when I do this problem. $$\int \frac{\mathrm{d}x}{x\sqrt{49x^2-1}}$$ Can someone please help me figure out what the correct answer should be?

• This does not look like a definite integral... – Igor Rivin Dec 5 '13 at 22:07

Hint: Substitute $u=\sqrt{49x^2-1}$

It's not a simple one but also not that hard...:

$$u:=\sqrt{49x^2-1}\;\;,\;\;du=\frac{49x}{\sqrt{49x^2-1}}dx\implies \frac{dx}x=\frac{u}{49x^2}du=\frac{u}{u^2+1}du\implies$$

$$\int\frac{dx}{x\sqrt{49x^2-1}}=\int\frac{u\,du}{u^2+1}\frac1{u}=\int\frac{du}{1+u^2}=\arctan u+C=\ldots$$

I would go with $49x^2-1=(7x)^2-1$. Then the standard trig substitution is $7x=\sec\theta$.

For $x>0$ $$\int \frac{\mathrm{d}x}{x\sqrt{49x^2-1}}= \int \frac{\mathrm{d}x}{x^2\sqrt{49-(\frac{1}{x})^2}}=-\int \frac{\mathrm{d}u}{\sqrt{49-u^2}}=-\arcsin\frac{u}{7}= -\arcsin\frac{1}{7x}+C.$$