How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$

Question

Question:

Find the integral $$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$

my solution: let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$ then $$t(\sqrt{x+1}-\sqrt{x})=1$$ $$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$ $(1)-(2)$ we have $$2\sqrt{x}=t-\dfrac{1}{t}\Longrightarrow x=\dfrac{1}{4}(t-\dfrac{1}{t})^2$$ so $$dx=\dfrac{1}{2}(t-\dfrac{1}{t})(1+\dfrac{1}{t^2})dt=\dfrac{t^4-1}{2t^3}dt$$

$$I=\int\dfrac{1}{1+t}\cdot\dfrac{t^4-1}{2t^3}dt=\dfrac{1}{2}\int\left(1+\dfrac{1}{t}+\dfrac{1}{t^2}+\dfrac{1}{t^3}\right)dt=\dfrac{1}{2}\left(t+\ln{t}-\dfrac{1}{t}-\dfrac{1}{2t^2}+C\right)$$ so $$I=\dfrac{1}{2}\left(\sqrt{x}+\sqrt{x+1}+\ln{(\sqrt{x}+\sqrt{x+1})}-\dfrac{1}{\sqrt{x}+\sqrt{x+1}}-\dfrac{1}{2(\sqrt{x}+\sqrt{x+1})^2}+C\right)$$

My question: have other methods? Thank you very much

Answer 1

since \begin{align*}I&=\int\dfrac{\sqrt{x}+1-\sqrt{x+1}}{(\sqrt{x}+1)^2-(\sqrt{x+1})^2}dx=\int\dfrac{\sqrt{x}+1-\sqrt{x+1}}{2\sqrt{x}}dx=\dfrac{1}{2}\int\left(1+\dfrac{1}{\sqrt{x}}-\sqrt{1+\dfrac{1}{x}}\right)dx\\ &=\dfrac{1}{2}\left(x+2\sqrt{x}-\int\sqrt{1+\dfrac{1}{x}}dx\right) \end{align*} since $$\int\sqrt{1+\dfrac{1}{x}}dx=x\sqrt{1+\dfrac{1}{x}}+\dfrac{1}{2}\ln{\left(2\left(\sqrt{1+\dfrac{1}{x}}+1\right)x+1\right)}+C$$

Answer 2

How about: substitute $x=u^2.$ Then (dropping absolute values for now), you get

$$\int \frac{2 u}{1+u+ \sqrt{u^2+1}} du$$

Now, substitute $u=\tan \theta,$ to get

$$2 \int \frac{\tan \theta \sec^2 \theta}{1+\tan \theta + \sec \theta} d\theta = \int \frac{\sin \theta}{\cos^2\theta (\sin \theta + \cos \theta + 1)} d\theta.$$

Now, make the substitution $t = \tan \frac{\theta}2,$ and you have a rational function.

Answer 3

Or, after the first substitution in my other answer, substitute $u=\sinh \theta.$ Then the integral becomes:

$$2\int \frac{\sinh \theta \cosh \theta}{1+\sinh \theta +\cosh \theta} d \theta = \int \frac{\sinh 2 \theta}{1 + e^\theta} d \theta.$$

The last integrand is a rational function of $e^\theta,$ so the integral obviously reduces to a rational function integral.

Answer 4

Using $\sqrt{x}=u=\tan(\theta)$ and $v=\sin(\theta)$, $$ \begin{align} &\int\frac1{1+\sqrt{x}+\sqrt{x+1}}\,\mathrm{d}x\\ &=\int\frac{1+\sqrt{x}-\sqrt{x+1}}{2\sqrt{x}}\,\mathrm{d}x\\ &=\sqrt{x}+\frac x2-\int\frac{\sqrt{x+1}}{2\sqrt{x}}\,\mathrm{d}x\\ &=\sqrt{x}+\frac x2-\int\sqrt{u^2+1}\,\mathrm{d}u\\ &=\sqrt{x}+\frac x2-u\sqrt{u^2+1}+\int\frac{u^2}{\sqrt{u^2+1}}\,\mathrm{d}u\\ &=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\int\tan^2(\theta)\sec(\theta)\,\mathrm{d}\theta\\ &=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\int\frac{\sin^2(\theta)}{\cos^4(\theta)}\,\mathrm{d}\sin(\theta)\\ &=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\int\frac{v^2}{1-2v^2+v^4}\,\mathrm{d}v\\ &=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\frac14\int\left(\frac1{(1-v)^2}+\frac1{(1+v)^2}-\frac1{1-v}-\frac1{1+v}\right)\,\mathrm{d}v\\ &=\sqrt{x}+\frac x2-\sqrt{x^2+x}-\frac14\log\left(\frac{1+v}{1-v}\right)+\frac14\frac1{1-v}-\frac14\frac1{1+v}+C\\ &=\sqrt{x}+\frac x2-\sqrt{x^2+x}-\frac12\log\left(\frac{1+\sin(\theta)}{\cos(\theta)}\right)+\frac12\frac{\sin(\theta)}{\cos^2(\theta)}+C\\ &=\sqrt{x}+\frac x2-\frac12\sqrt{x^2+x}-\frac12\log(\sqrt{x}+\sqrt{x+1})+C \end{align} $$ since $\tan(\theta)=\sqrt{x}$ and $\sec(\theta)=\sqrt{x+1}$