Solve an integral by bringing quadratic trinomial to canonical form

Question

The problem is the following: $$\int{\sqrt{4x^2+x}dx}$$ Now once gotten to canonical form of a quadratic trinomial, $ax^2+bx+c=a(x-(\frac{-b}{2x}))^2-\frac{b^2-4ac}{4a}$, the intregral looks like this:

$$\int{\sqrt{4(x+\frac{1}{8})^2-\frac{1}{16}}dx}=\int{\sqrt{(2x+\frac{1}{4})^2-\frac{1}{16}}dx}$$ With the substition of $2x+\frac{1}{4}=t$, $2dx=dt\implies dx=\frac{dt}{2}$ integral becomes: $$\int{\sqrt{t^2-\frac{1}{16}}dx}$$ But what to do from here. I've seen a few solutions that involve a lot of trigonometry, but the book from which I got this problem states that the solution they got is: $$\frac{2+x}{2}\sqrt{4x^2+x}-2ln|x+2+\sqrt{4x^2+x}|$$ How did they get to this exact solution?

Answer 1

Let $s=8x+1.$

$\int\sqrt{4x^2+x}\,dx=\frac1{32}\int\sqrt{s^2-1}\,ds$

Integrating by parts,

$$\begin{align}\int\sqrt{s^2-1}\,ds&=[s\sqrt{s^2-1}]-\int\frac{s^2}{\sqrt{s^2-1}}\,ds\\&=[s\sqrt{s^2-1}]-\int\frac{s^2-1}{\sqrt{s^2-1}}\,ds-\int\frac1{\sqrt{s^2-1}}\,ds \end{align}$$ hence $$\int\sqrt{s^2-1}\,ds=\frac12\left([s\sqrt{s^2-1}]-\int\frac1{\sqrt{s^2-1}}\,ds\right).$$ Now (V p. 263 of your book) $$\int\frac1{\sqrt{s^2-1}}\,ds=[\ln|s+\sqrt{s^2-1}|]$$ and we can conclude: $$\begin{align}\int\sqrt{4x^2+x}\,dx&=\frac1{64}[s\sqrt{s^2-1}-\ln|s+\sqrt{s^2-1}|]\\&=\frac1{64}[4(8x+1)\sqrt{4x^2+x}-\ln|8x+1+4\sqrt{4x^2+x}|] \\&=\left[\frac{8x+1}{16}\sqrt{4x^2+x}-\frac1{64}\ln\left|\frac{8x+1}4+\sqrt{4x^2+x}\right|\right]. \end{align}$$ (WolframAlpha agrees.)

(N.B. our notation "$[\dots]$" means "$\dots+$ constant", and in the two last expressions, the constants differ.)

This seems to be the intended method of your book but its solution is false, since it does not differ by a constant from this one.

Answer 2

Substitute $t=x+a$, along with $a=\frac18$ $$I = \int \sqrt{4x^2+x}\ dx=2\int \sqrt{t^2-a^2}\ dt $$

and integrate by parts $$I = t \sqrt{t^2-a^2} -a^2\int \frac{dt}{\sqrt{t^2-a^2}} $$ To evaluate the remaining integral, let $s={\sqrt{t^2-a^2}}$ to get $sds=tdt$

$$\int \frac{dt}{\sqrt{t^2-a^2}}=\int \frac{d(t+s)}{t+s}=\ln|t+s| $$

As a result \begin{align} I= t \sqrt{t^2-a^2} -a^2\ln|t + \sqrt{t^2-a^2}|\\ \end{align}