Question

sorry lag.... integration using substitution

Answer 1

\[\int\limits_{}^{}x \sqrt{\left( x+3 \right)}dx\]

Answer 2

Let \(u=x+3\) => du/1
\(x = u-3\)

Answer 3

\[\int\limits u-3\sqrt{u} du\]
Then integrate it..

Answer 4

you should say\[\int\limits_{}^{}(u-3)\sqrt{u}du\]

Answer 5

letting u know im still here working on it lol

Answer 6

woops, sorry forgot the brackets..

Answer 7

the way i do this..let u = sqrt x - 3
u^2 = x-3
2udu = dx

x = u^2 -3 so....


(u^2-3)(u)(2udu)

pull out the 2... int (u^2)(u^2 - 3)du

int (u^4 - 3u^2)du

use the power rule thing...

this is my method...whichever you think is easier ^_^

Answer 8

yes usually I'll let u^2=something because of square root

Answer 9

but looks more complicated..

Answer 10

dont forget to substitute back to terms of x after integrating and distribute 2 :D

Answer 11

im getting there

Answer 12

im confused

Answer 13

why so? which method did you use? mines or igb's?

Answer 14

i tried both

Answer 15

Well, where did you get stuck?

Answer 16

sorry guys im making it more difficult than it is

Answer 17

ok..... u= x+3 and du = dx ????

Answer 18

i dont know how you get x-3

Answer 19

\[\int\limits_{}^{}x \sqrt{x+3}dx\]
Let
-----------------------------------
u^2=x+3 -----> | u^2-3=x
Differentiate |
2udu=dx |
-------------------------------------
Sub into equation,
\[\int\limits_{}^{}x \sqrt{x+3}dx\]
\[\int\limits\limits_{}^{}(u^2-3)u(2udu)\]
\[2\int\limits\limits\limits_{}^{}(u^2-3)u^2du\]
\[2\int\limits\limits\limits_{}^{}u^4-3u^2du\]
\[2[\frac{u^5}{5}-\frac{3u^3}{3}]+C\]
\[2[\frac{u^5}{5}-u^3]+C\]
\[2[\frac{(x+3)^{5/2}}{5}-(x+3)^{3/2}]+C\]

Answer 20

ohhhh