Question

what is the integral of ((1+sqrt.x)^1/3)/(sqrt.x)

Answer 1

wolframalpha will save u, use it:
wolframalpha.com

Answer 2

http://www.wolframalpha.com/input/?i=ntegral+of+%28%281%2Bsqrt%28x%29%5E1%2F3%29%2F%28sqrt%28x%29

Answer 3

I got 2/3*(1+sqrt.x)^3/2

Answer 4

click on show steps

Answer 5

that not the right problem

Answer 6

no i think not. if you want to cheat you have to be careful

Answer 7

i dont want to cheat, i worked it out and got 2/3*(1+sqrt.x)^3/2

Answer 8

rather simple u-sub yes?

\[\int\frac{\sqrt[3]{1+\sqrt{x}}}{\sqrt{x}}dx\]

Answer 9

right

Answer 10

u=1+sqrt.x

Answer 11

put
\[u=1+\sqrt{x}\]
\[du=\frac{1}{2\sqrt{x}}\] get
\[\frac{1}{2}\int\sqrt[3]{u}du\]

Answer 12

get
\[\frac{1}{2}\times \frac{3}{4}u^{\frac{4}{3}}\]
\[=\frac{3}{8}\sqrt[3]{(1+\sqrt{x}})^4\]

Answer 13

oh no mistake!!

Answer 14

sorry the constant out front should be 2, not 1/2

Answer 15

\[du=\frac{1}{2\sqrt{x}}dx\]
\[2du=\frac{dx}{\sqrt{x}}\]

Answer 16

so instead of
\[\frac{3}{8}\] it should be
\[\frac{3}{2}\] out front. my mistake

Answer 17

oh its raised to 4/3

Answer 18

dont tell me the answer yet

Answer 19

ok i let you finish it

Answer 20

i got it it is :

Answer 21

close but you are off by a constant.

Answer 22

+ C

Answer 23

should be
\[\frac{3}{2}\] not
\[\frac{3}{4}\]

Answer 24

no i meant the constant out front. because when you do the u-sub you do not get
\[du=\frac{1}{\sqrt{x}}dx\] you get
\[du=\frac{1}{2\sqrt{x}}dx\]

Answer 25

so you get
\[2du=\frac{1}{\sqrt{x}}dx\] pull the 2 out front

Answer 26

that is why it should be
\[\frac{3}{2}\] not
\[\frac{3}{4}\]

Answer 27

but i took the derivative of the answer i got and it gave the integral

Answer 28

no. the derivative of
\[\sqrt{x}=\frac{1}{2\sqrt{x}}\]

Answer 29

so you will be off by a factor of 2

Answer 30

\[\frac{d}{dx}\frac{3}{4}(1+\sqrt{x})^{\frac{4}{3}}\]
\[=\frac{4}{3}\times \frac{3}{4}(1+\sqrt{x})^{\frac{1}{3}}\times \frac{1}{2\sqrt{3}}\]

Answer 31

so with out the extra two out front you have the wrong answer. but easy to fix, just multiply by 2

Answer 32

http://www.twiddla.com/577354

Answer 33

can you cheack my work

Answer 34

i am there