# $\int \sqrt{\frac{x}{x+1}}dx$

In order to integrate

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

I did:

$$x = \tan^2\theta$$

$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$

$$p = \cos\theta \implies dp = -\sin\theta d\theta$$

$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta = -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp = -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$

$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$

$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$

But the result seems a little bit different than wolfram alpha. I Know there may be easier ways to solve this integral but my question is about this method I choose, specifically.

Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan$ to something simpler?

• It looks like things started falling apart with the square root. You never took the square root of the denominator. And the next step doesn't appear to follow. – Mike Feb 10 '15 at 21:55
• $\sqrt{\tan^2(\theta)+1}=\sqrt{\sec^2(\theta)}=|\sec^2(\theta)|$ Also – kingW3 Feb 10 '15 at 21:57
• @Mike :'( $\frac{}{}$ – Guerlando OCs Feb 10 '15 at 22:03
• @kingW3 $\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=|\sec\theta|$ – Mike Feb 10 '15 at 22:10
• @Mike That was supposed to be the point of my comment about the mistake,I guess I was too hasty – kingW3 Feb 10 '15 at 22:18

For the last question, you have a triangle with side lengths $1,x,\sqrt{x^2+1}$ to have tangent $x$. That triangle has cosine $\frac{1}{x^2+1}$. Hence $$\cos \arctan x=\frac{1}{\sqrt{x^2+1}}$$

Let $u=\sqrt{\frac{x}{x+1}}$. Then \begin{align} \int\sqrt{\frac{x}{x+1}}\,\mathrm{d}x &=\int u\,\mathrm{d}\frac{u^2}{1-u^2}\\ &=\frac12\int u\,\mathrm{d}\left(\frac1{1-u}+\frac1{1+u}\right)\\ &=\frac12\int u\left(\frac1{(1-u)^2}-\frac1{(1+u)^2}\right)\,\mathrm{d}u\\ &=\frac12\int\left(\frac1{(1-u)^2}-\frac1{1-u}+\frac1{(1+u)^2}-\frac1{1+u}\right)\,\mathrm{d}u\\ &=\frac12\left(\frac1{1-u}+\log(1-u)-\frac1{1+u}-\log(1+u)\right)+C\\ &=\frac{u}{1-u^2}+\frac12\log\left(\frac{1-u}{1+u}\right)+C\\[6pt] &=\sqrt{x(x+1)}+\log\left(\sqrt{x+1}-\sqrt{x}\right)+C \end{align}

Consider $$I=\int\sqrt{\frac{x}{1+x}}dx.$$ We proced by the change of variable $u=\sqrt{1+x}$, $du=\frac{1}{2\sqrt{1+x}}$dx and $x=u^2-1,$ which gives $$I=2\int \sqrt{u^2-1}\ du=u\sqrt{u^2-1}-\log\left(u+\sqrt{u^2-1}\right)+C,$$ see https://owlcation.com/stem/How-to-Integrate-Sqrtx2-1-and-Sqrt1-x2 for more informations. The calculations ends by replacing value of u by $\sqrt{x^2-1}$.

Another aproach to the integral : substitution $x=t^2$

$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \int\frac{2t^2 \mathrm{d}t}{\sqrt{t^2+1}}$$

By per partes :

$$I = \int\frac{t^2 \mathrm{d}t}{\sqrt{t^2+1}} = t\sqrt{t^2+1}-\int\sqrt{t^2+1}\;\mathrm{d}t = t\sqrt{t^2+1}-\int\frac{t^2+1}{\sqrt{t^2+1}}\;\mathrm{d}t$$

Ergo $$I = t\sqrt{t^2+1}-I-\operatorname{arcsinh}{t}$$ So, because original integral was $2I$, solution is therefore :

$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \sqrt{x}\sqrt{x+1}-\operatorname{arcsinh}{\sqrt{x}} +C$$