Question

Hello!! Is anyone good with evaluating indefinite integrals?

Answer 1

This one seems easy but I could sure use confirmation on the correct answer!! Lol.

Answer 2

post the question

Answer 3

Posting!!

Answer 4

uh uhs, amistre is here :P

Answer 5

.......TRIG......

Answer 6

i meant uh ohs*

Answer 7

I couldn't figure out how to post a division with out the sign!

Answer 8

give me a T; give me an R; give me an IG...whats that spell? trouble?

Answer 9

lols nah, spells awesomeness

Answer 10

post it your best and well decipher it :)

Answer 11

\frac{numer}{deno}

Answer 12

ArcsinX + C

Answer 13

^-- in the equation editor
and like amistre said it's not a big deal

Answer 14

lols all yours amistre

Answer 15

integrate arcsin + C? or is that your answer that yougot?

Answer 16

That's my answer - I just have to evaluate the integral

Answer 17

whats the integral so that we can pervue its intricate innards :)

Answer 18

\[\frac{1}{\sqrt{1 - x^2}}\]
is this the integral?

Answer 19

without the integral sign...or the dx...

Answer 20

\frac\[\int\limits_{?}^{?}\frac{1}{1+x^2} dx\]

Answer 21

thats a arctan..

Answer 22

yup, amistre's right

Answer 23

i saw it in a movie lol

Answer 24

lols "down the rabbit hole"?

Answer 25

:) tan^1(x) + C...are there any bounds to evaluate?

Answer 26

..^-1..

Answer 27

\[[\int\limits] dx 1 + x2 = \arctan x + C \]

Answer 28

Was that all I needed to refer to?! I have it in my notes but the 'dx' was in the num

Answer 29

its like learning greek all over again ;)

Answer 30

something like this?
\[\int\limits_{}^{}\frac{1}{x^2 + 1}dx\]

Answer 31

Noooo kidding! lol

Answer 32

Noooo kidding! lol

Answer 33

\[\int\limits_{} \frac{1}{1+x^2} dx \rightarrow \tan^{-1}(x) + C\]

Answer 34

for answering this question, amistre...i give you...a MEDAL

Answer 35

-applause-

Answer 36

yay!! ..... its prolly foiled covered chocolate...so I accept :)

Answer 37

lols how'd you guess?
enjoy!

cheers

Answer 38

Will arctan x + c be correct?

Answer 39

-nod-

Answer 40

as long as you havent left anyting out, yes; but arctan is a dated term

Answer 41

i would know; im dated ;)

Answer 42

lol!

Answer 43

Can you guys help me with another problem/confirm my answer being right or wrong?!!

Answer 44

perhaps, but display the question first; just makes it easier :)

Answer 45

lol - kk!!

Answer 46

Eval the Indefinite Integral :\[\int\limits\limits_{?}^{?}\frac{x^3 dx}{\sqrt[4]{5x^4-1}}\]

Answer 47

you want a 15 for that :)

Answer 48

I got: 1/15 (5x^4)^3/4 + C

Answer 49

its refered to as "u- substitution"

u = 5x^4 - 1; du = 20x^3 dx

dx = du/20x^3

Answer 50

\[\frac{1}{15} (5x^{4})^{3/4} + C\]

Answer 51

\[\int\limits_{} \frac{x^3 }{20x^3 u^{1/4}}du\]

Answer 52

i mistyped 15; meant 20 :)

the x^3s cancel; pull out the 1/20; and integrate the u^(-1/4)

Answer 53

\[\frac{1}{20} \int\limits_{} u^{-1/4} du \rightarrow \frac{1}{20} \frac{u^{(-1+ 4)/4}}{(-1+4)/4}\]

Answer 54

u^(3/4)/(3/4)

4 * 4root(u^3)
------------
20 * 3


4root(u^3)
------------ = 4root[5x^4-1]/15
5 * 3

Answer 55

forgot to ^3 my (5x^4 - 1)

Answer 56

\[\frac{\sqrt[4]{(5x^4 -1)^3}}{15} +C\]

Answer 57

You are great!!!!! So final answer: \[\frac{1}{20} (5x^{4} -1)^{3/4} + C?\]

Answer 58

that looks comparabley good :)

Answer 59

ack!!...1/15.....not 1/20

Answer 60

ahh!!

Answer 61

Fantastic!! lol - I almost messed that one up! Thank you so much! I have another problem I might post a new question so I can award medals - lol

Answer 62

im sure we will be around :)