Question

hello guys, here somewhat challenging substitution integral.
\[\int \frac{\sqrt[4]{x}}{\sqrt{x}+\sqrt[3]{x}}dx\]

Answer 1

is it an addition of x^1/3 's on the bottom?

Answer 2

oh sorry the bottom is sqrt(x)+root3(x)

Answer 3

i corrected it

Answer 4

this was in the u sub section

Answer 5

i tried u=x^1/12
and then i had to do another sub along the lines

Answer 6

somehow i arrived at this expression \[\frac{4}{3}x^{1/2}-4x^{1/6}+\arctan(x^{1/2})+c\]

Answer 7

but it is not right i must of did something wrong along the process

Answer 8

http://www.wolframalpha.com/input/?i=int%28%28x%5E1%2F4%29%2F%28x%5E1%2F2%2Bx%5E1%2F3%29%29dx
this is wolfram's answer

Answer 9

My process was this, \[\Large x= u^{12} \\ \Large dx = 12 u^{11} du\]

Answer 10

eh that's pretty close to what i did @Kainui
i must of just missed the detail perhaps

Answer 11

Let's see, I didn't finish but it looks like this will take us where we need to go.

Answer 12

yes seems so i did u=x^1/12
which is the same as x=u^12

Answer 13

\[\Large \int\limits \frac{u^{12/4}}{u^{12/2}+u^{12/3}} 12 u^{11}du = \\ \Large 12 \int\limits \frac{u^{10}}{1+u^2}du\]

Answer 14

hmm i guess that's where i missed it
i got a different expression, darn it lol

Answer 15

my proccess was this
\(\Huge\sf{^P_B{^e_u{^a_t{^n_t{^u_e{^t_r}}}}}\hspace{-65pt}\LaTeX}\)

Answer 16

sorry... urge was to strong... *runs for the hills*

Answer 17

Yeah, maybe something with exponent rules? \[\Large u=\tan \theta \\ \Large du = \sec^2 \theta d \theta\] \[\Large 12 \int\limits \tan^{10} \theta d \theta\]

Answer 18

yes i checked it i made a mistake with exponents

Answer 19

for me i would do long division for that one?

Answer 20

instead of trig sub?

Answer 21

hmm does not seem a good thing to do long division haha

Answer 22

seems that i got \[12\int( u^8-u^6+u^4-u^2-\frac{1}{1+u^2})du\]

Answer 23

Yeah I mean whatever you gotta do to get the answer haha.

Answer 24

thanks guys
i will check my steps and see :)

Answer 25

what about my solution? you said whatever i got to do to get the answer... xD

Answer 26

by the way how would you deal with tan^10x
seems to me it has a lot of work itself

Answer 27

Well you're right about that, I just know there are reduction formulas you could use. I'm not really saying that it's pretty haha.

Answer 28

haha yeah it is not pretty at all
seems what i got it correct above i just completed the rest