OpenStudy (anonymous):

integrate this:

OpenStudy (anonymous):

$\int\limits_{16}^{1} \frac{ 1 }{ \sqrt{x} (1+ \sqrt[4]{x})} dx$

OpenStudy (anonymous):

Please reverse the 1 and 16.

OpenStudy (anonymous):

@amistre64 @ParthKohli

OpenStudy (anonymous):

@mathmale

OpenStudy (kirbykirby):

I would try doing this: substitute $$\large y=x^{1/4}$$ this means $$\large y^2=x^{1/2}$$ Then you have the integral: $\frac{1}{y^2(1+y)}$ Then use partial fractions to get: $\frac{1}{y^2} +\frac{1}{y+1}-\frac{1}{y}$ and this is easy to integrate. Just make sure you re-substitute back into your original x-values!

OpenStudy (anonymous):

where did you put the du problem though.

OpenStudy (anonymous):

OpenStudy (anonymous):

Whenever you substitute, don't forget to adjust with the derivative.

OpenStudy (anonymous):

OpenStudy (kirbykirby):

Oh i just used the y as a placeholder here just to reformulate the question to get "easy to work with integers" so that I could do the partial fractions easier... you just have to substitute the appropriate x values into the "partial-fractioned" integrand: $\frac{1}{x^{1/2}}+\frac{1}{x^{1/4}+1}-\frac{1}{x^{1/4}}$

OpenStudy (anonymous):

Oh i see. Yep you are right!

OpenStudy (anonymous):

I would replace 4th root of x by u, adjust the whole thing to get u/(u+1) du and solve. Would that take less time?

OpenStudy (anonymous):

By the way, integrating 1/(4throot(x)+1) is really long right? I can't just say that it is ln(4throot(x)+1)?

OpenStudy (anonymous):

(I'm in Cal II, I have my final, but I have not done integration for at least 1.5 months! The machine is rusty.)

OpenStudy (kirbykirby):

oh right it is not trivial to do that integration.. hmm let me think a little on this!

OpenStudy (anonymous):

OpenStudy (kirbykirby):

integrating 1/(4throot(x)+1)

OpenStudy (kirbykirby):

Oh wait maybe a trig substitution would work

OpenStudy (anonymous):

did it work?

OpenStudy (kirbykirby):

Um actually I went with another method which is working much better! (I'll write it below)

OpenStudy (kirbykirby):

Use $$\large x^{1/4}=u$$ $$\large du=\frac{1}{4}x^{-3/4}\implies dx = 4x^{3/4}du$$ This gives: $\int\frac{1}{x^{1/2}(u+1)}4x^{3/4}du=\int\frac{4x^{1/4}}{u+1}du=4\int\frac{u}{u+1}du$ Partial fraction, or eye-balling, shows that $\frac{u}{u+1}=1-\frac{1}{u+1}$ so, $4\int\left(1-\frac{1}{u+1} \right)du=4(u-\log(u+1))+C$

OpenStudy (kirbykirby):

sorry for my initial method.. when I used the y's, it looked easy to integrate, but I didn't really realize that putting back in the x's made it more difficult actually