Question

Need help solving an integral...

Answer 1

post it..

Answer 2

\[\int\limits_{0}^{2} 2t/(t-3)^2dt\]

Answer 3

sorry the limits of integration are 0 to 2

Answer 4

think you use partial fractions for this one

Answer 5

Have you done partial fractions mdntjem?

Answer 6

I drive nascars

Answer 7

No let u = t-3
then t = u+3
Then integral is 2(u+3)/u^2

Answer 8

u = t-3
du = dt
t = u+3
Integral ( 2(u+3)/u^2 du )

Answer 9

Now multiply the top out:
2(u+3)/u^2 = (2u + 6)/u^2 = [2u/u^2] + [6/u^2]

Answer 10

\[\frac{2t}{(t-3)^2}=\frac{2}{t-3}+\frac{6}{(t-3)^2}\]

Answer 11

i think i heart math way is easier.

Answer 12

Cancel out one u and you'll end up with:
2/u + 6/u^2
Integrate:
2 ln|u| + 6 [u^(-2+1)]/[-2+1]

Answer 13

I mean, either way, partial fractions still would have worked :P

Answer 14

Yea the partials would work, but I think u - sub is easier..
2Ln|u| - 6/u + C

Answer 15

Now just plug u back in:
2Ln|t-3| - 6/(t-3) and evaluate..

Answer 16

you are right. partial fractions a pain, but the form looked like a set up for them. in any case you have the answer as
\[2\ln(|t-3|)-\frac{6}{t-3}\]

Answer 17

for a "final answer" of
\[-2\ln(3)+4\]

Answer 18

{2Ln|2-3| - 6/(2-3)} - {2Ln|0-3| - 6/(0-3)}=
{2Ln1 - 6/-1} - {2Ln3 - 6/-3} = {0 + 6} - 2Ln3 - 2 = 4 - 2Ln3 I confirm satellite's answer.

Answer 19

Partial fractions isn't too bad in this case though. You only need to solve 2 constants. Nice and simple. It seems to me though, that method^^ (writing u in terms of t and then replugging in) is the most under-used method. Nothing wrong with it, but I feel as though most people would try partials.

Answer 20

Actually malevolence, I used to use partials in every single integral... then I found out I would save a lot more time if I used u substitution..

Answer 21

lol

Answer 22

My teachers kept telling me to use u substitution, partial fractions is like a harsh yet effective way to solve.. like brute force method..

Answer 23

I mean, I'm not downing your method. I'm just saying when most people learn u-sub they learn it for stuff like:
\[\int\limits 2xe^{x^2}dx\]
Where the du is plain and in sight.
Most people (that I know of) would think to go if u=2x+1 then x=u/2-1/2.
Or something like that. See what I mean?

Answer 24

\begin{array}l\color{red}{\normalsize\text{p}}\color{orange}{\normalsize\text{a}}\color{#9c9a2e}{\normalsize\text{r}}\color{green}{\normalsize\text{t}}\color{blue}{\normalsize\text{i}}\color{purple}{\normalsize\text{a}}\color{purple}{\normalsize\text{l}}\color{red}{\normalsize\text{ }}\color{orange}{\normalsize\text{f}}\color{#9c9a2e}{\normalsize\text{r}}\color{green}{\normalsize\text{a}}\color{blue}{\normalsize\text{c}}\color{purple}{\normalsize\text{t}}\color{purple}{\normalsize\text{i}}\color{red}{\normalsize\text{o}}\color{orange}{\normalsize\text{n}}\color{#9c9a2e}{\normalsize\text{s}}\color{green}{\normalsize\text{ }}\color{blue}{\normalsize\text{a}}\color{purple}{\normalsize\text{r}}\color{purple}{\normalsize\text{e}}\color{red}{\normalsize\text{ }}\color{orange}{\normalsize\text{a}}\color{#9c9a2e}{\normalsize\text{ }}\color{green}{\normalsize\text{p}}\color{blue}{\normalsize\text{a}}\color{purple}{\normalsize\text{i}}\color{purple}{\normalsize\text{n}}\color{red}{\normalsize\text{}}\end{array}

Answer 25

looool

Answer 26

yea I know.. and I also know that people tend not to use the u-sub if they see fractions.. but i think hell twist your brain cells and figure out a shortcut.. but it's just me haha..

Answer 27

satellite do you know linear algebra?

Answer 28

I love I love math

Answer 29

in fact all of these integration are a pain. just math teachers showing off. look i can integrate this, and i can integrate that. fact is if you picked a function out of a hat the probability it is integrable is 0. in real life unfortunately you need numeric methods.

Answer 30

True. That's why I'm glad I'm done with calculus. Going into proof based classes this fall :)

Answer 31

yea lol.. true satellite.. my professor gave us an extra credit problem once... but actually integrals are my favourite topic!

Answer 32

I was attempting to do it with partial fractions, but was not getting anywhere

Answer 33

i used to know some linear algebra. in fact last night i spent maybe an hour helping someone do some nice gram schmidt orthogonalization. talk about a pain

Answer 34

Yeah, it wasn't really that hard mdntjem

Answer 35

mdntjem, it's all about practice.. I've solved like hundreds of them haha.. not tryin to show off but after hours and hours of lookin up answers and methods, you look at the problem and know which method will work the fastest.. anyway satellite can you go on my question..spanning sets?

Answer 36

Here:
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e077c300b8bd74af49bef7a

Answer 37

I'm trying to get there :) Gotta keep doing a crap ton of these integrals. Calc 2 for summer semester goes wicked fast tho.

Answer 38

thank you all for the help!

Answer 39

mdntjem if you want to do partial fractions you can do:
\[\int\limits \frac{2t}{(t-3)^2}dt=\int\limits \frac{A}{t-3}+\frac{B}{(t-3)^2}dt\]
Get a common denominator and set the numerators equal.
A(t-3)+B=2t
At-3A+B=2t
That means A=2 and -3A+B=0
So: -3(2)+B=0
B=6
So you have:
\[\int\limits \frac{2}{t-3}dt+\int\limits \frac{6}{(t-3)^2}dt\]
Giving:
\[2\ln|t-3|-\frac{6}{t-3}+C\]
Then you can evaluate it.

Answer 40

@bahrom7893 I recommend you http://faculty.luther.edu/~macdonal/LAGAintro.pdf

Answer 41

estudier, will take a look.. thanks..

Answer 42

Estudier, that is only the cover, intro, and index for me :(

Answer 43

Here it is at Amazon
http://www.amazon.com/Linear-Geometric-Algebra-Alan-Macdonald/dp/1453854932

He has a paper that is somewhat similar to the book, I will find it and post link.

Answer 44

http://faculty.luther.edu/~macdonal/GA&GC.pdf

This concentrates rather more on the GA part rather than the LA part whereas the book integrates the two for undergraduates.

If you know LA very well, the connection is obvious.