2
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ This does not look like a definite integral... $\endgroup$ – Igor Rivin Dec 5 '13 at 22:07
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.