Integral of $\sqrt{ {\rm ln}^2 4 \cdot 4^{2 x} + 1}$

Question

I'm currently taking calculus, and have hit a problem that is causing me confusion. I have the answer to the problem, I just have no idea how to arrive at that answer. The problem is as follows:

$$\int\sqrt{\ln^2\left(4\right)\space\cdot\space 4^{2x}+1}\mathrm dx$$

I assume you have to use $u$-substition, but I don't know what to use as $u$. I am very stumped on what to do.
Technically it is a definite integral, but once I figure out the integration, I'm confident I can evaluate over the ranges.

Thank you in advance to anyone who can help me.

Answer 1

First you need to forget about all these numbers - they are here to confuse you. Use the following:

$$4^{2x}=16^x=e^{\ln 16 \cdot x}$$

Now denote:

$$a=\ln^2 4$$

$$b=\ln 16$$

Now your integral becomes:

$$\int \sqrt{a~e^{bx}+1}~~dx$$

Now you need to change the variable, for example (in the comments there is another great substitution):

$$u=\sqrt{a~e^{bx}+1}$$

$$x=\frac{1}{b} \ln \frac{u^2-1}{a}$$

$$dx=\frac{1}{b} \frac{2u }{u^2-1} du$$

$$\int \sqrt{a~e^{bx}+1}~~dx=\frac{2}{b} \int \frac{u^2 }{u^2-1} ~du=$$

$$=\frac{2}{b} \int \frac{u^2-1}{u^2-1} ~du+\frac{2}{b} \int \frac{1}{u^2-1} ~du=$$

$$=\frac{2u}{b}+\frac{2}{b} \int \frac{du}{u^2-1} $$

The last integral I leave to you

Answer 2

Just use the simplest substitution ever:

$$y = \sqrt{\ln^2(4)\cdot 4^{2x} + 1} ~~~~~~~ \text{d}y = \frac{1}{2\sqrt{\ln^2(4)\cdot 4^{2x} + 1}}\cdot (2\cdot \ln^2(4)\cdot \ln(4)\cdot 4^{2x}) = \frac{\ln(4)(y^2-1)}{y}\ \text{d}x$$

Thus you get

$$\ln(4)\int\frac{y^2}{y^2-1}\ \text{d}y$$

Which can be computed by doing a split of fraction:

$$\frac{y^2}{y^2-1} = -\frac{1}{2(y+1)} + \frac{1}{2(y - 1)} + 1$$

In this way you split into three integrals, the last of which is trivial.

You can do these alone.