Question

Integrate sqrtx / sqrtx + 1

Answer 1

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

Answer 2

I'ved tried substitution, but it gets more complex

Answer 3

multiply and divide by root x-1

Answer 4

sorry?

Answer 5

rationalize the denominator

Answer 6

\[\int\limits_{}^{}\frac{ \sqrt{x} }{ \sqrt{x} + 1 } * \frac{ \sqrt{x} + 1 }{ \sqrt{x} + 1 } dx\]

Answer 7

yup

Answer 8

hmm,ok

Answer 9

teacher never showed us this technique

Answer 10

then just do substitution afterwards?

Answer 11

it gets more complex

Answer 12

oh, wait...that won't work....

Answer 13

put u = root x +1

Answer 14

oo maybe that'll work

Answer 15

then root x = u-1

Answer 16

dx= 2(u-1) du

Answer 17

and then it becomes easy to solve ..... :)

Answer 18

ask if any doubts or if u didn't get it...

Answer 19

ok, ill post my answer in a sec

Answer 20

\[\int\limits_{}^{}2u + \int\limits_{}^{}\frac{ 2 }{ u }-\int\limits_{}^{}4 \]

Answer 21

yes, going good....proceed.

Answer 22

is the antiderivative of 4, 4x or 4u

Answer 23

4u cause im in terms of u right now

Answer 24

yes, 4u

Answer 25

\[u^2+u-4u\]

\[(\sqrt{x}+1)^2+(\sqrt{x}+1)-4(\sqrt{x}+1)\]

Answer 26

for 2/u , its 2log u

Answer 27

oh yea..

Answer 28

\[(\sqrt{x}+1)^2+2\ln(\sqrt{x}+1)-4(\sqrt{x}+1)\]

Answer 29

then expand

Answer 30

correct.

Answer 31

ok thanks hartnn!

Answer 32

welcome ^_^