Evaluate the Integral $\int\sqrt{\frac{2x+1}{3x+2}}dx$

Question

Evaluate the integral:$$\int\sqrt{\frac{2x+1}{3x+2}}dx$$

My approach:

$$y:=\sqrt{3x+2}\Rightarrow dy=\frac{3}{2\sqrt{3x+2}}\\\therefore x=\frac{y^2-2}{3}$$

Rewriting the integral in terms of $y$:

$$\int\sqrt{\frac{2x+1}{3x+2}}dx{ =\frac{2}{3}\int\sqrt{\frac{2y^2-1}{3}}dy\\ =\frac{2\sqrt2}{3\sqrt3}\int\sqrt{y^2-\left(\frac{1}{\sqrt2}\right)^2}dy\\ =\frac{2\sqrt2}{3\sqrt3}\left(\frac{y}2\sqrt{y^2-\left(\frac{1}{\sqrt2}\right)^2}-\frac{1}{4}\ln\left|y+\sqrt{y^2-\left(\frac{1}{\sqrt2}\right)^2}\right|\right)+c\\ =\frac{2\sqrt2}{3\sqrt3}\left(\frac{\sqrt{(3x+2)(2x+1)}\sqrt3}{2\sqrt2}-\frac{1}{4}\ln\left|\sqrt{3x+2}+\sqrt{3x+\frac{3}2}\right|\right)+c\\ =\frac{1}3\sqrt{(3x+2)(2x+1)}-\frac{1}{3\sqrt{6}}\ln\left|\sqrt{3x+2}+\sqrt{3x+\frac{3}2}\right|+c}$$

But , answer in the book is: $$\frac{1}3\sqrt{(3x+2)(2x+1)}-\frac{1}{3\sqrt{6}}\ln\left|\sqrt{2x+1}+\sqrt{2x+\frac{4}3}\right|+c$$

It is an indefinite integral so answer may defer. But, it dosen't look the same.

Answer 1

$3x+\frac{3}{2}=\frac{3}{2}(2x+1)$ and $3x+2=\frac{3}{2} \left ( 2x + \frac{4}{3} \right )$, so the arguments of the two logs differ by a constant factor (namely $\sqrt{3/2}$), meaning that the logs themselves differ by a constant (namely $\ln(\sqrt{3/2})$). Since everything else is the same, the results without the $+c$'s differ by a constant. So these answers are equivalent.

Answer 2

$\sqrt[n]{\dfrac{a\,x+b}{c\,x+d}}$ With a common denominator of powers $\left(\dfrac{2\,x+1}{3\,x+2}\right)^{\mathbf{n}}$ with $\mathbf{n}=\dfrac{1}{2}$ and $p=2$

$$\def\arraystretch{2}\begin{array}{c|c}u=\dfrac{\sqrt{2\,x+1}}{\sqrt{3\,x+2}}&x=\dfrac{1-2\,{u}^{2}}{3\,{u}^{2}-2}\\&\mathrm{d}x=\dfrac{2\,u}{\left({3\,{u}^{2}-2}\right)^{2}}\,\mathrm{d}u\end{array}$$ $$\int{\dfrac{2\,{u}^{2}}{\left({3\,{u}^{2}-2}\right)^{2}}}{\;\mathrm{d}u}$$ Integration by parts $\int{fg'}=fg-\int{f'g}$ $$\def\arraystretch{2}\begin{array}{c|c}f=u&g'=\dfrac{u}{\left({3\,{u}^{2}-2}\right)^{2}}\\f'=1&g=\,{\displaystyle\int{\dfrac{u}{\left({3\,{u}^{2}-2}\right)^{2}}\;\mathrm{d}u}=\dfrac{1}{6}\,\int{\dfrac{1}{\left({3\,{u}^{2}-2}\right)^{2}}\;\mathrm{d}\left(3\,{u}^{2}-2\right)}=-\dfrac{1}{6\,\left(3\,{u}^{2}-2\right)}}\end{array}$$ $$2\left(-\dfrac{u}{6\,\left(3\,{u}^{2}-2\right)}+\int{\dfrac{1}{6\,\left(3\,{u}^{2}-2\right)}}{\;\mathrm{d}u}\right)=$$ =-$$\dfrac{\ln\left(\left|\sqrt{3}\,u+\sqrt{2}\right|\right)}{2\,\sqrt{2}\,3\,\sqrt{3}}+\dfrac{\ln\left(\left|\sqrt{3}\,u-\sqrt{2}\right|\right)}{2\,\sqrt{2}\,3\,\sqrt{3}}-\dfrac{u}{3\,\left(3\,{u}^{2}-2\right)}$$

Notes

$$\int{\dfrac{1}{6\,\left(3\,{u}^{2}-2\right)}}{\;\mathrm{d}u}$$ $$\dfrac{1}{6}\int{\dfrac{1}{3\,\left({u}^{2}-\dfrac{2}{3}\right)}}{\;\mathrm{d}u}$$

$$\dfrac{1}{18}\int{\dfrac{1}{\left(u-\dfrac{\sqrt{2}}{\sqrt{3}}\right)\,\left(u+\dfrac{\sqrt{2}}{\sqrt{3}}\right)}}{\;\mathrm{d}u}$$

Grouping $$1=-\dfrac{\sqrt{3}}{2\,\sqrt{2}}\,\left(\left(u-\dfrac{\sqrt{2}}{\sqrt{3}}\right)-\left(u+\dfrac{\sqrt{2}}{\sqrt{3}}\right)\right)$$ $$\dfrac{1}{18}\int{-\dfrac{\sqrt{3}\,\left(\dfrac{1}{u+\dfrac{\sqrt{2}}{\sqrt{3}}}-\dfrac{1}{u-\dfrac{\sqrt{2}}{\sqrt{3}}}\right)}{2\,\sqrt{2}}}{\;\mathrm{d}u}$$ $$\dfrac{1}{18}\left(-\dfrac{\sqrt{3}}{2\,\sqrt{2}}\int{\dfrac{1}{u+\dfrac{\sqrt{2}}{\sqrt{3}}}}{\;\mathrm{d}u}+\dfrac{\sqrt{3}}{2\,\sqrt{2}}\int{\dfrac{1}{u-\dfrac{\sqrt{2}}{\sqrt{3}}}}{\;\mathrm{d}u}\right)=$$

$$=\dfrac{\ln\left(\left|\sqrt{3}\,u-\sqrt{2}\right|\right)}{4\,\sqrt{2}\,3\,\sqrt{3}}-\dfrac{\ln\left(\left|\sqrt{3}\,u+\sqrt{2}\right|\right)}{4\,\sqrt{2}\,3\,\sqrt{3}}$$

$$\int{\dfrac{1}{u+\dfrac{\sqrt{2}}{\sqrt{3}}}}{\;\mathrm{d}u}$$ $$\int{\dfrac{\sqrt{3}}{\sqrt{3}\,u+\sqrt{2}}}{\;\mathrm{d}u}$$ $$\def\arraystretch{2}\begin{array}{c|c}v=\sqrt{3}\,u+\sqrt{2}&u=\dfrac{v-\sqrt{2}} {\sqrt{3}}\\&\mathrm{d}u=\dfrac{1}{\sqrt{3}}\,\mathrm{d}v\end{array}$$ $$\sqrt{3}\int{\dfrac{1}{\sqrt{3}\,v}}{\;\mathrm{d}v}$$

Substitution $$\begin{array}{c|c}v=\sqrt{3}\,u+\sqrt{2}&u=\dfrac{v-\sqrt{2}}{\sqrt{3}}\end{array}$$

$$\ln\left(\left|\sqrt{3}\,u+\sqrt{2}\right|\right)$$