Evaluate $\int \frac{dx}{x\sqrt{4x^2+1}}$

Question

I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals.

My work

I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals. I found in the Table of Integrals an integral that is akin to $\int \frac{dx}{x\sqrt{4x^2+1}}$. The integral I was talking about is $$\int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a} ln \left( \frac{a+\sqrt{u^2+a^2}}{u} \right)$$

With that in mind, I need to modify $\int \frac{dx}{x\sqrt{4x^2+1}}$ to look like $\int \frac{du}{u\sqrt{u^2+a^2}}$.

Modifying now: $$\int \frac{dx}{x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}}$$

The modified integral to be evaluated would be $\int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}}$.

Now getting the integral:

$$\int \frac{dx}{x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = -2\left(\left(\frac{1}{(1)}\right) ln \left( \frac{(1)+\sqrt{(2x)^2+(1)^2}}{(2x)} \right) \right)$$ $$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = -2 ln \left( \frac{1+\sqrt{4x^2+1}}{2x} \right) $$ $$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = ln \left(\left( \frac{1+\sqrt{4x^2+1}}{2x} \right)^{-2} \right)$$ $$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = \frac{1}{ln \left(\left( \frac{1+\sqrt{4x^2+1}}{2x} \right)^2 \right)}$$ $$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = ln \left( \frac{4x^2}{1+\sqrt{4x^2+1}}\right)$$

So...the integral of $\frac{dx}{x\sqrt{4x^2+1}}$ would be $ln \left( \frac{4x^2}{1+\sqrt{4x^2+1}}\right)$. But in the book I used, it is $ln \left( \frac{x}{1+\sqrt{4x^2+1}}\right)$. It was so close to my answer, yet I don't know where I messed up.

Where did I messed up?

There are many problems in your work. Please review the log rules. — Simply Beautiful Art, Oct 01 '17 at 15:39
You're using the substitution $u = 2x$... What is $du$ in terms of $dx$? — cip999, Oct 01 '17 at 15:39
After the first equal sign after "Now getting the integral: ..." you have a missing $2$ in the denominator. — Simply Beautiful Art, Oct 01 '17 at 15:41
@SimplyBeautifulArt Table of Integrals is a list of common integrals.....It was usually printed at the end of the book.....so it's a handy reference.....The problem asks me to evaluate the integral above using a Table of Integral", so I followed..... — fitzmerl duron, Oct 01 '17 at 15:43
Hm, okay. I've never gotten my hands on a calculus textbook per se, so I wouldn't know what such a thing were @PalautotKa — Simply Beautiful Art, Oct 01 '17 at 15:44
@SimplyBeautifulArt You can find a sample of "Table of Integrals" in the Internet to know what it look like.....and I edited my post...referring to "Now getting the integral: ..." you have a missing 2 in the denominator. " — fitzmerl duron, Oct 01 '17 at 15:47
Hm, okay. I usually know the basic integrals off the top of my head or ask WolframAlpha for quick things. — Simply Beautiful Art, Oct 01 '17 at 15:51

score 2 · Accepted Answer · answered Oct 01 '17 at 15:53

2

After "Now getting the integral", when you use the formula, you introduce a $2$ that should not be there. You would get $$ -\ln\left(\frac{1+\sqrt{4x^2+1}}{x}\right) =\ln\left(\frac{x}{1+\sqrt{4x^2+1}}\right). $$ You also need to review the rules for logarithms. It is not true that $\ln 1/x=1/\ln x$ (you used it twice, so it actually "cancelled itself", but it is stil wrong).

answered Oct 01 '17 at 15:53

Martin Argerami

205,756

I thought $\frac{1}{x\sqrt{4x^2+1}}$ would be the same as $\frac{2}{2x\sqrt{4x^2+1}}$ or $\frac{2}{2x\sqrt{(2x)^2+(1)^2}}$ – fitzmerl duron Oct 01 '17 at 16:00
Exactly, so the two $2$ you need cancel each other, and you don't need to introduce a $2$ in the formula. – Martin Argerami Oct 01 '17 at 16:01
Oh I get it now! That "$2$"! Thanks for helping me out, man! I get it when you said "you don't need to introduce a 2 in the formula."...... – fitzmerl duron Oct 01 '17 at 16:03
Wait...what about this operation? $\int \frac{dx}{x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = -\left(\frac{1}{(1)}\right) ln \left( \frac{(1)+\sqrt{(2x)^2+(1)^2}}{(2x)} \right)$.......I was using this integral $\int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a} ln \left( \frac{a+\sqrt{u^2+a^2}}{u} \right)$.......Why is it that it becomes $\ln\left(\frac{x}{1+\sqrt{4x^2+1}}\right)$ instead of $ ln \left( \frac{2x}{1+\sqrt{4x^2+1}}\right)$? – fitzmerl duron Oct 02 '17 at 03:22
You missed the minus sign. – Martin Argerami Oct 02 '17 at 03:25
Where's the minus sign? What part of the operation? – fitzmerl duron Oct 02 '17 at 03:31
1

Yeah, sorry I was thinking about something else. That $2$ makes no difference. The antiderivative is not unique. Both $\ln\left(\frac{x}{1+\sqrt{4x^2+1}}\right)$ and $\ln\left(\frac{2x}{1+\sqrt{4x^2+1}}\right)$ are valid antiderivatives, they differ by a constant. – Martin Argerami Oct 02 '17 at 03:34

Evaluate $\int \frac{dx}{x\sqrt{4x^2+1}}$

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