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

Question

How to evaluate $$\int\sqrt{\frac{5+x}{7+x}}dx$$? I am trying this question by rationalizing the denominator. So, the integral will become $$\int{\frac{\sqrt{{(5+x)}×{(7+x)}}}{7+x}}dx$$. Now I am thinking for substituting $x$ $=$ $7×(tan\theta)^{2}$. So, $dx$ $=$ $14×tan\theta×(sec\theta)^{2}×d\theta$. So, finally the integral will be $$\int\frac{\sqrt{{(5+7(tan\theta)^{2})}×{7(sec\theta)^{2}}}}{(sec\theta)^{2}}×2×tan\theta×(sec\theta)^{2}×d\theta$$. 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

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 4

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

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