Question

integral of 1/(sqrt(x)+cube root of x)

Answer 1

\[\int \frac{\text{d}x}{\sqrt{x}+\sqrt[3]{x}}\]right?

Answer 2

yes. Sorry..I didnt know how to write it another way

Answer 3

no problem :)

Answer 4

have u tried it yet?

Answer 5

the problem says to make a substitution to express the integrand as a rational function and then avaluate the integral.

Answer 6

I have, but am not sure how to go about with the substitution

Answer 7

using substitution \[x=t^6\]

Answer 8

the book gave a hint and is saying to use u=\[\sqrt[6]{x?}\]

Answer 9

but i dont know how they got that.

Answer 10

or where to plug it in

Answer 11

yes its same\[u=\sqrt[6]{x}\]\[u^6=x\]u use that sub to get rid of radicals

Answer 12

perfect instruction, hand off, @mukushla

Answer 13

oh ok thanks

Answer 14

so would it be 1/(u^6+u^1/2) ?

Answer 15

thanks @Loser66

looking more carefully\[\int \frac{6t^5\text{d}t}{t^3+t^2}=\int \frac{6t^3\text{d}t}{t+1}\]

Answer 16

sorry, its better to work with ur notation\[\int \frac{6u^5\text{d}u}{u^3+u^2}=\int \frac{6u^3\text{d}u}{u+1}\]

Answer 17

if you have x=u^6, then wouldnt \[\sqrt[3]{x}\] with x^6 inside be x^1/2?

Answer 18

u^6 will be inside not x^6 :)

Answer 19

oh ok

Answer 20

im not sure how you got the u^2 on the denominator

Answer 21

well\[\sqrt[3]{x}=\sqrt[3]{u^6}=(u^6)^{\frac{1}{3}}=u^{\frac{6}{3}}=u^2\]

Answer 22

makes sense? :)

Answer 23

oh i see it now..thanks alot!

Answer 24

anytime :)

Answer 25

and you got the 6u^5 by taking the derivative of u right?

Answer 26

yes.

Answer 27

\[\int \frac{6u^3\text{d}u}{u+1}\] step up from this by long division and very easy to get the answer then, right?

Answer 28

yes i did long division and then used partial fractions

Answer 29

what do you get from long division?

Answer 30

u^2-u+1-(1/u+1)

Answer 31

yep, for the last one, no need to use partial, you have formula for this, =ln|u+1|

Answer 32

don't forget, replace u by x at the final answer

Answer 33

i did. Thanks alot!

Answer 34

give medal to mukusla, he gave you a perfect instruction. beautiful substitution. !!!