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

$\sqrt[n]{\dfrac{a\,x+b}{c\,x+d}}$ 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 $$\begin{array}{c|c}v=\sqrt{3}\,u+\sqrt{2}&u=\dfrac{v-\sqrt{2}}{\sqrt{3}}\end{array}$$

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

Answer 2

$3x+\frac{3}{2}=\frac{3}{2}(2x+1)$ 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 3

Another approach is to rationalize the numerator:

$$\int\sqrt{\frac{2x+1}{3x+2}}\mathrm dx$$ $$=\int\frac{2x+1}{\sqrt{6x^2+7x+2}}\mathrm dx$$

Rewrite the numerator of the integrand as:$$2x+1=\frac16(12x+7)-\frac16$$

Answer 4

The exponential substitution offers a simple and general approach for any integral of the form $\int f\left(x, \frac{\sqrt{x+m}}{\sqrt{x+n}}\right)\,dx$ since integrals of this type always reduce to the integral $\frac{-ai}{2} \int (e^{-i\alpha} + e^{i\alpha} - 2)\, d\alpha$, where $\alpha = \cos^{-1}\left(\frac{x+b}{a}\right)$. Here, here, and here you can see more examples solved by this method.

Let $b-a=m$ and $b+a=n$, where $a$ and $b$ are real numbers. Then the following identities hold:

$$\frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}=\tan{\left(\frac12\sec^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{\sqrt{x+b-a}}{\sqrt{x+b+a}}=\frac{\sqrt{x+m}}{\sqrt{x+n}},\tag{1}$$

$$e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b - \sqrt{(x + b)^2 - a^2}}{a}.\tag{2}$$

These identities are generalizations of identities $(1-2)$ and $(9-10)$ at "Integration using Euler-like identities" and can be derived similarly as shown in the linked blog. The general identities $(1)$ and $(2)$ lead us to the following general transformation formula:

$$\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{3}$$ Where $\alpha=\cos^{-1}\left(\frac{x+b}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x+b}{a}\right)$ and $\gamma=\sec^{-1}\left(\frac{x+b}{a}\right).$ Use the upper sign for the alternating signs $\mp$ when $\frac{x + b}{a} \geq 1$ and the lower sign when $0 \leq \frac{x + b}{a} \leq 1$.

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Solution. First, let's rewrite the integral as follows $$\int \sqrt{\frac{2x+1}{3x+2}}\,dx=\sqrt{\frac{2}{3}}\int \frac{\sqrt{x+\frac12}}{\sqrt{x+\frac23}}\,dx.$$

Now we obtain the values of $a$ and $b$ by solving the following (simple!) system of equations:

$$\left. b-a=\frac12\atop a+b=\frac23\right\}$$

The solutions are $a = \frac{1}{12}$ and $b = \frac{7}{12}$. Applying formula $(3)$ for $x\geq-\frac12$, the integral becomes

$$\begin{aligned}\int \frac{\sqrt{2x+1}}{\sqrt{3x+2}}\,dx&= -\frac{i\sqrt{6}}{72}\int \frac{(1 - e^{i\alpha}) (e^{-i\alpha} - e^{i\alpha})}{(e^{i \alpha} + 1)} \, d\alpha\\&=-\frac{i\sqrt{6}}{72}\int \frac{e^{-i\alpha} (e^{i\alpha} - 1) (e^{2i\alpha} - 1)}{e^{i\alpha} + 1} \, dx\\&=-\frac{i\sqrt{6}}{72}\int e^{-i\alpha}(e^{i\alpha}-1)^2\,d\alpha\\&=-\frac{i\sqrt{6}}{72}\int (e^{-i\alpha}+e^{i\alpha}-2)\,d\alpha\\&=\frac{\sqrt{6}}{72}(2i\alpha+e^{-i\alpha}-e^{i\alpha})+C\end{aligned}$$

To switch to real numbers, plug in the values of $a$ and $b$ into $(2)$ and simplify. Then, replace the expression in the antiderivative from $(2)$ and simplify again. Thus, we obtain

$$=\frac{1}{36} \left( \sqrt{6} \ln \left| 12x - 2\sqrt{6} \sqrt{6x^{2} + 7x + 2} + 7 \right| + 12 \sqrt{6x^{2} + 7x + 2} \right) + C $$