How to evaluate $\int\sqrt{\frac{5+x}{7+x}}dx$?

Question

How to evaluate $$\int\sqrt{\frac{5+x}{7+x}}\mathrm dx \, ?$$ I am trying this question by rationalizing the denominator. So, the integral will become $$\int{\frac{\sqrt{{(5+x)}\times {(7+x)}}}{7+x}}\mathrm dx.$$ Now I am thinking for substituting $x = 7 \times (\tan\theta)^{2}$. So, $\mathrm dx = 14 \times \tan\theta \times (\sec\theta)^{2}\times \mathrm d\theta$. So, finally the integral will be $$\int\frac{\sqrt{{(5+7(\tan\theta)^{2})}\times{7(\sec\theta)^{2}}}}{(\sec\theta)^{2}}\times 2 \times \tan\theta \times (\sec\theta)^{2} \times \mathrm d\theta.$$ But I can't approach further. Please help me out.

Answer 1

$\textbf{Hint:}$ Use the substitution $x+5 = 2\sinh^2 t$

$$I = \int \sqrt{\frac{2\sinh^2t}{2+2\sinh^2t}}4\sinh t \cosh t dt = \int 4\sinh^2tdt$$

then use the double angle identity $\cosh^2 t + \sinh^2 t = 1 + 2\sinh^2 t = \cosh 2t $. Can you complete the computation?

Answer 2

Hint

$$I=\int\sqrt{\frac{5+x}{7+x}}\,dx$$

Get rid of the radical using $$\sqrt{\frac{5+x}{7+x}}=t \implies x=\frac{5-7 t^2}{t^2-1}\implies dx= ???\,dt$$ and you will get a simple expression.

Use partial fraction decomposition to face very simple and standard integrals.

Answer 3

A general approach for integrals of the form $\int f\left(x,\frac{\sqrt{x+m}}{\sqrt{x+n}}\right)\,dx$

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 $0 \leq \frac{x + b}{a} \leq 1$.

Solution 1. First, we obtain the values of $a$ and $b$ by solving the following (simple!) system of equations:

$$\left. b-a=5\atop a+b=7\right\}$$

The solutions are $a = 1$ and $b = 6$. Applying formula $(3)$ for $x\geq-5$, the integral becomes

$$\begin{aligned}\int \frac{\sqrt{x+5}}{\sqrt{x+7}}\,dx&= -\frac{i}{2}\int \frac{(1 - e^{i\alpha}) (e^{-i\alpha} - e^{i\alpha})}{(e^{i \alpha} + 1)} \, d\alpha\\&=-\frac{i}{2}\int \frac{e^{-i\alpha} (e^{i\alpha} - 1) (e^{2i\alpha} - 1)}{e^{i\alpha} + 1} \, dx\\&=-\frac{i}{2}\int e^{-i\alpha}(e^{i\alpha}-1)^2\,d\alpha\\&=-\frac{i}{2}\int (e^{-i\alpha}+e^{i\alpha}-2)\,d\alpha\\&=\frac12(2i\alpha+e^{-i\alpha}-e^{i\alpha})+C\end{aligned}$$

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

$$=\sqrt{x^2+12x+35} + \ln \left| x - \sqrt{x^2+12x+35} + 6 \right|+C$$

Now suppose you have to evaluate an integral like this one:

$$\int \frac{x\sqrt{x+1}}{\sqrt{x+2}}\,dx$$

Traditional approaches can become very complicated (as you can see in the linked calculator), whereas with this method, you only need a few lines of algebra and basic calculus like in the case of your integral.

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Solution 2. This substitution is dual, so I’d better start thinking of a better name than Exponential Substitution. Indeed, a cleaner solution is possible using the following extended Weierstrass substitution for irrational integrands: $$\int f\left( x, \sqrt{(x + b)^2 - a^2}, \dfrac{ \sqrt{x + b - a} }{ \sqrt{x + b + a} } \right) \, dx= \int f\left(\frac{t^2+1}{2t}a - b, \frac{t^2-1}{2t}a, \frac{1-t}{1+t}\right) \frac{t^2-1}{2 t^2}a\, dt,\tag{4}$$ where $t=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b - \sqrt{(x + b)^2 - a^2}}{a}$, for $\frac{x+b}{a}\geq1$. Thus, $$\begin{aligned}\int\sqrt{\frac{x+5}{x+7}}\,dx&=\int\frac{1-t}{1+t}\cdot\frac{t^2-1}{2t^2}\,dt\\&=-\int\frac{t^2-2t+1}{2t^2}\,dt\\&=-\frac12\int(1-2t^{-1}+t^{-2})\,dt\\&=\ln\left|t\right| +\frac12\left(\frac{1}{t}-t\right)+C\end{aligned}$$ To express the antiderivative in terms of $x$, you can use $(2)$ as in my previous solution.

The transformation formula $(4)$ seems to incorporate one or more Euler substitutions, since when solving certain integrals and comparing both techniques, I have noticed that we arrive at the same rational expression. However, to me, applying $(4)$ feels more intuitive and straightforward.

It would seem that there is no longer a need to use complex exponentials, but for the interval $[-1, 1]$, they might still be useful since the inverse cosecant is not defined for an argument equal to zero.

Answer 4

Let $\displaystyle y=\sqrt{\frac{5+x}{7+x}}$, then $$ x=\frac{7 y^2-5}{1-y^2} \textrm{ and } d x=\frac{4 y^2}{\left(1-y^2\right)^2} d y $$ The integral was converted into an integral of rational function and can be found by integration by parts. $$ \begin{aligned} & I=\int y \cdot \frac{4 y}{\left(1-y^2\right)^2} d y \\ & =2 \int y d\left(\frac{1}{1-y^2}\right) \\ & =\frac{2 y}{1-y^2}-2 \int \frac{d y}{1-y^2} \\ & =\frac{y}{1-y^2}+\ln \left|\frac{1-y}{1+y}\right|+C \\ & =(7+x) \sqrt{\frac{5+x}{7+x}}+\ln \left|\frac{1-\sqrt{\frac{5+x}{7+x}}}{1+\sqrt{\frac{5+x}{7+x}}}\right|+C \end{aligned} $$

Answer 5

$$I:=\int \sqrt{\frac{5+x}{7+x}} dx = I:=\int \sqrt{1-\frac{2}{7+x}} dx$$

let $\frac{2}{7+x} =y $ then $dx= \frac{-2dy}{y^2}$

$$I= -2 \int \frac{\sqrt{1-y} }{y^2} dy $$ let $y=\cos^2(t)$ then $dy =-2 \sin(t) \cos(t)$ and $\sec(t) =\frac{1}{\sqrt{y}} = \sqrt{\frac{2}{7+x}} $ and $\sin(y)=\sqrt{1-y^2} =\sqrt{1-\left(\frac{2}{7+x}\right)^2}$

$$I=4\int \frac{\sin(t)}{\cos^4(t)}\sin(t) \cos(t)dt =4\int \frac{\sin^2(t)}{\cos^3(t)}dt =4 \int \sec^3{x} -\sec( x)dt= \frac{1}{2} (-\ln|\sec(x)+ \tan(x) | +\sec(x) \tan(x)) +C $$ see how to evaluate $\int \sec^3(t)dt$ here $$= \frac{1}{2}\left(-\ln\left|\sqrt{\frac{2}{7+x}} +\sqrt{\frac{2}{7+x}}\sqrt{1-\left(\frac{2}{7+x}\right)^2} \right| +\sqrt{1-\left(\frac{2}{7+x}\right)^2} \frac{2}{7+x} \right)+C $$