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

I tried with $u=\sqrt x $, but this did not work. I really don't know what to do...

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    $\begingroup$ You look like you're on the right track with your substitution. Where are you stymied from there? $\endgroup$ – Steven Stadnicki Oct 22 '14 at 23:53
  • $\begingroup$ Well it becomes $\frac{\sqrt{1+u}}{u^2}$ and dx=2udu right? But what about (i don't know the translation so what about a and b if i have $\int_a^b$ ? but there are not here..) ? :( $\endgroup$ – ParaH2 Oct 22 '14 at 23:59
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Let $u=\sqrt{1+\sqrt{x}}$ to get $x=(u^2-1)^2$ and $dx=4u(u^2-1)du$, so

$\displaystyle\int\frac{\sqrt{1+\sqrt{x}}}{x}dx=\int\frac{4u^2}{u^2-1}du=4\int\left(1+\frac{1}{u^2-1}\right)du$

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It looks like you should be able to eliminate both problem radicals with the substitution

$$x=\tan^4t,dx=4\tan^3t\sec^2tdt$$ $$\int\dfrac{\sqrt{1+\sqrt{x}}}{x}dx=\int\frac{\sec t}{\tan^4 t}4\tan^3t\sec^2tdt=$$ $$\int\frac{4\sec^3tdt}{\tan t}=4\int\frac{dt}{\sin t\cos^2t}=4\int\frac{\sin tdt}{\cos^2t(1-\cos^2t)}$$ $$u=\cos t,du=-\sin tdt$$ $$-4\int\frac{du}{u^2(1-u^2)}=-4\int\frac{du}{u^2}-4\int\frac{du}{1-u^2}=$$ $$-4\int\frac{du}{u^2}-2\int\frac{du}{1-u}-2\int\frac{du}{1+u}$$ So far, it's been relatively straightforward, but back-substitution may be an issue. We have

$$\frac4u+2\ln(1-u)-2\ln(1+u)=4\sec t+2\ln\frac{1-\cos t}{1+\cos t}=$$ $$4\sec t+2\ln\frac{\sec t-1}{\sec t+1}$$

From here, you'd have to use $\sec t=\sqrt{1+\sqrt x}$. You could clean up the numerator of that natural log by rewriting it as $\ln\dfrac{\tan^4t}{(\sec t+1)^4}$, but that denominator is still going to be ugly...

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  • $\begingroup$ Oh gosh ^^ good luck for the end of your calculus, for me there is may be an arcth in the answer but i don't know :/ $\endgroup$ – ParaH2 Oct 23 '14 at 2:16

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