Question

Integral question...

Answer 1

well i have one way in mind
let sqrt(x)=u
x=u^2
and partial fractions
-
though i might try to look for an easier way

Answer 2

I haven't learned partial fractions yet-:(

Answer 3

then try the sub I mentioned

Answer 4

and then see if you can see another sub to go from there

Answer 5

this can be done without partial fractions

Answer 6

and I don't have that derivative of u in there.

Answer 7

\[u=\sqrt{x} \\ u^2=x \\ 2 u du=dx\]

Answer 8

you can do this sub
and then another
or you could combine the subs:
\[1+\sqrt{x}=u \\ \sqrt{x}=u-1 \\ x=(u-1)^2 \\ dx=2(u-1) du\]

Answer 9

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

Answer 10

So you are just re-writing the derivative of U, not in terms of x, but in terms of u, so that we can substitute it into the integral, right?

Answer 11

\[\int\limits_{ }^{ }\frac{2u}{(1+u)}du\]so this is right?

Answer 12

Anyone?

Answer 13

that is to the 4th power on bottom

and i'm having trouble on your questions
so
instead of differentiating u=sqrt(x) w.r.t. x
I prefer to differentiate u^2=x w.rt. x

Answer 14

but you can differentiate either and still win

Answer 15

\[\int\limits_{ }^{ }\frac{2u}{(1+u)^4}du\]

Answer 16

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

Answer 17

yes, forgot the 4th power.

Answer 18

and what sub will you choose after that one?

Answer 19

\[\int\limits_{ }^{ }\frac{2u}{(1+u)^4}+\frac{2}{(1+u)^4}-\frac{2}{(1+u)^4}du\]
\[\int\limits_{ }^{ }\frac{2u+2}{(1+u)^4}-\frac{2}{(1+u)^4}du\]\[2\int\limits_{ }^{ }\frac{1}{(1+u)^3}du~-~2\int\limits_{ }^{ }\frac{1}{(1+u)^4}du\]

and from there it is easy even for me....

Answer 20

a very unusual and perhaps silly approach, but looks simple and not so much calculus-wise.

Answer 21

that is interesting to look at it that way...
I like it

Answer 22

\[-\frac{1}{(1+u)^2}+\frac{2}{3(1+u)^3}+C\]

Answer 23

am I right?

Answer 24

and then sub in sqrt x for u.

Answer 25

looks good

Answer 26

\[\frac{1}{3(1+\sqrt{x})^3}-\frac{1}{(1+\sqrt{x})^2}+C\]

Answer 27

\[\int\limits_{}^{}\frac{1}{(1+\sqrt{x})^4} dx \\ 1+\sqrt{x}=u \\ u-1=\sqrt{x} \\ (u-1)^2=x \\ 2(u-1)du=dx \\ \int\limits_{}^{}\frac{2(u-1) du}{u^4}=\int\limits_{}^{}(2u^{-3}-2 u^{-4})du \\ =\frac{2u^{-2}}{-2}-2 \frac{u^{-3}}{-3}+C \\ =\frac{-1}{u^2}+\frac{2}{3u^3}+C \]
\[=\frac{-1}{(1+\sqrt{x})^2}+\frac{2}{3( 1+\sqrt{x})^3}+C\]

Answer 28

you are missing your 2 somewhere

Answer 29

i submitted the way I was doing it in just case you wanted to look at it

Answer 30

but the way you chose is really awesome

Answer 31

yes, I am missing a 2 on top of the fraction where (sqrtx+1) is to the third power, but that doesn't mater once I understand how to play with u substitution the way you did it....

Answer 32

ty for the help:)

Answer 33

np