Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Voting to close this down since other methods are going to be only some slight variants of the solution posted by OP. –  user17762 Dec 7 '13 at 4:14
    
No.I think this problem maybe have other methods.Thank you –  china math Dec 7 '13 at 4:20

4 Answers 4

up vote 3 down vote accepted

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 $$\int\sqrt{1+\dfrac{1}{x}}dx=x\sqrt{1+\dfrac{1}{x}}+\dfrac{1}{2}\ln{\left(2\left(\sqrt{1+\dfrac{1}{x}}+1\right)x+1\right)}+C$$

share|improve this answer

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.

share|improve this answer

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 $\sec(\theta)=\sqrt{x+1}$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.