Question

help please 13 , 17 , 21

Answer 1

\[\rm \int\limits_{}^{} \frac{ dx }{ 5-3x }\]
is same as \[\large\rm \int\limits_{ }^{ } \frac{ 1 }{ 5-3x } dx\]

Answer 2

oh okay so then

Answer 3

the goal is to get rid of the x from the denominator
are you familiar with u-sub ?

Answer 4

yes , i just had no clue what the teacher was saying lol

Answer 5

D:

Answer 6

opps sorry ii fell asleep...

u-sub is the easy way to deal with complicated integrals
\[\int\limits_{ }^{} g(x) \color{Red}{dx} \]
let u =g(x)
take the derivative
\[\rm du=g'(x) dx\]
solve for dx\[ \color{blue}{\frac{du}{g'(x)}}=dx\]
replace dx with du/g'(x)
\[\int\limits_{ }^{} g(x) \color{red}{\frac{du}{g'(x)}} \]

Answer 7

oh is there a formula for this? lol

Answer 8

would this be it ?

http://www.sosmath.com/calculus/integration/rational/Tablefraction/img1.gif

Answer 9

no that's an example.
that is pretty easy (&fun) the difficult part is to figure out `u`
you have to play with the function to find out but usually for composite function u will be the inner function

Answer 10

wohow what's that lemme see

Answer 11

okie lol i found it online D: but does that always work ?

Answer 12

no seriously what's that ??
is that for u-substitution ?

Answer 13

hmm i think so

Answer 14

i have no idea sorry
i don't think there are formulas for this .. just the example i've posted
some questions are really easy where you just have to take anti-derivative to integrate
and for some you have to use u-sub method

Answer 15

oh okay can you guide me through the steps of this problem D:

Answer 16

\[\large\rm \int\limits_{ }^{ } \frac{ 1 }{ 5-3x } \color{Red}{dx}\]like for this question
we can't split this fraction(and we to get rid of that x variable from the denominator )
let \[\large\rm u =5-3x \]
now take the derivative

Answer 17

oh so we just let the bottom = 5-3x for u ?

Answer 18

yes you will see the reason at the end...

Answer 19

oh okay so then itll be -3

Answer 20

\[\large\rm u=5-3x\]
Differentiate with respect to x
\[\large\rm \frac{du}{dx}= -3\]
solve for dx
\[\rm \frac{1}{-3}du= \color{Red}{dx}\]
now we can replace dx with (-1/3du) in original integral and we can replace `5-3x` with ul \[\large\rm \int\limits_{ }^{ } \frac{ 1 }{u } \cdot \color{Red}{\frac{du}{-3}}\]

Answer 21

now you can easily find the anti-derivative

Answer 22

hmm which one do i find the anti deriv?

Answer 23

\[\large\rm \int\limits_{ }^{ } \frac{ 1 }{u } \cdot \color{Red}{\frac{du}{-3}}\]
this one we don't need the original one anymore (it's hard to find the anti-derivative of the original integral which involves fraction(that's why we have to apply u-sub method))

Answer 24

so then itll be like
1 -1
_______ * _______ du
5-3x -3

Answer 25

u=5-3x so we can substitute (5-3x) for u
\[\int\limits_{ }^{ } \frac{ 1 }{ u } \frac{-1}{3} du\]

Answer 26

1 1
_______ * - _______ du
5-3x 3

Answer 27

replace `5-3x` with u

Answer 28

thats the bottom part right ?

Answer 29

yes.

Answer 30

oh okay thats what u did or did i do something wrong ?

Answer 31

no that's correct
we don't want 5-3x at the denominator so that's why we assumed that u =5-3x