Question

integration problem

Answer 1

\[\int\limits_{0}^{1} (x ^{2}+3)^{-1} dx ^{-1}\]

Answer 2

no -1 after dx sorry

Answer 3

\[\int_0^1\frac{1}{x^2+3}dx\]?

Answer 4

yup

Answer 5

partial fraction

Answer 6

haven't learnt that yet sorry

Answer 7

hmmm

Answer 8

partial fractions wont work unless you are working in the complex plane using a contour, because
\[x^2+3\] is prime

Answer 9

look in the back of your text for integrals of the form
\[\frac{1}{u^2+a^2}\]

Answer 10

i tried but cannot continue..

Answer 11

thats tan

Answer 12

you will see
\[\int\frac{1}{u^2+a^2}du=\frac{1}{a}\tan^{-1}(\frac{x}{a}) + C\]

Answer 13

so your "anti derivative" is
\[\frac{1}{\sqrt{3}}\tan^{-1}(\frac{x}{\sqrt{3}})\]

Answer 14

everything you need to know in calc 2, save series (if it is in your syllabus) can be found on the back four pages of your text

Answer 15

Everyone attacked the question.

Answer 16

we can derive the formula if you like, but it is a collosal waste of time do to it more than once

Answer 17

of course you are not done. you have to find
\[\frac{1}{\sqrt{3}}(\tan^{-1}(\frac{1}{\sqrt{3}})-\tan^{-1}(0))\]

Answer 18

i haven't learnt the formula you stated and i can't even find it from the paper, the paper just provides those compound angle formulae but the others

Answer 19

are you using a text?

Answer 20

what do you mean?

Answer 21

1/3 -x^2/9 +x^4/27 - ...
----------------
3+x^2 ) 1
(1+x^2/3)
----------
-x^2/3
(-x^2/3 -x^4/9)
-----------------
x^4/9
(x^4/9+x^6/27)
-------------
-x^6/27
......

\[\sum_{n=0}^{inf}\frac{(-1)^{n+1}}{3^{n+1}}x^{2n}\]
maybe

Answer 22

once again you need to know that
\[\frac{d}{dx}[\tan^{-1}(x)]=\frac{1}{x^2+1}\] and then make an adjustment for the 3