Question

integrate this:

Answer 1

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

Answer 2

Please reverse the 1 and 16.

Answer 3

@amistre64 @ParthKohli

Answer 4

@mathmale

Answer 5

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!

Answer 6

where did you put the du problem though.

Answer 7

where is your 1/4(x^(3/4))

Answer 8

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

Answer 9

do you see your mistake?

Answer 10

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:

\[\frac{1}{x^{1/2}}+\frac{1}{x^{1/4}+1}-\frac{1}{x^{1/4}} \]

Answer 11

Oh i see. Yep you are right!

Answer 12

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?

Answer 13

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

Answer 14

(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.)

Answer 15

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

Answer 16

About what?

Answer 17

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

Answer 18

Oh wait maybe a trig substitution would work

Answer 19

did it work?

Answer 20

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

Answer 21

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,
\[ 4\int\left(1-\frac{1}{u+1} \right)du=4(u-\log(u+1))+C\]

Answer 22

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