Question

another integral (it is just in order of the practice list, of ones I haven't done b4)

Answer 1

\[\int\limits_{ }^{ }x^3\sqrt{1-x^2}~dx\]\[u=1-x^2,~~~~~~~~~-\frac{1}{2}du=x~dx~~~~~~~~~x^2=1-u\]

Answer 2

\[\int\limits_{ }^{ }(1-u)\sqrt{u}~du\]

Answer 3

oh, that even I can do.

Answer 4

Try this one out for size:
\[\int\frac{dx}{(x+1)(x^2+1)(x^3+1)}\]

Answer 5

it becomes,
\[\frac{2}{3}u \sqrt{u}-\frac{2}{5}u^2\sqrt{u}+C\]

Answer 6

what? I am not good at integration by partial fraction yet. I learned integration by parts online with Khan and other tutorials, but this is a bit too much for me

Answer 7

Ah what about
\[\int x^{1/4}\ln x\,dx~~?\]

Answer 8

I will think about this one....:)

Answer 9

\[\frac{4}{5}x^{5/4}\ln x-\frac{4}{5}\int\limits_{ }^{ }x^{1/4}~dx\]

Answer 10

like this, and then the obvious part, right?

Answer 11

(The X's canceled)

Answer 12

okay, I will post the answer too

Answer 13

\[\frac{4}{5}x^{5/4} \ln x-\frac{16}{25}x^{5/4}+C\]

Answer 14

I had a typo

Answer 15

the power was 5/4 not 4/5

Answer 16

@SithsAndGiggles I take ur challenge

Answer 17

I did it no?

Answer 18

for the first problem

Answer 19

lol

Answer 20

and the sith's problem?

Answer 21

Yup, by parts is the easiest way for that one.

Answer 22

@TheLOL by all means :)

Answer 23

you guys can measure strengths, but Zarkon will be stronger jk

Answer 24

\[A/(x+1) + Bx+c/(x^2+1) + (dx^2 + e)/(x^3+1)\]

Answer 25

?

Answer 26

Or Expand x^3+1

Answer 27

hey I just turned ORANGE

Answer 28

you will be chaning score slower and slower

Answer 29

@fbi2015 feel free to browse these links for extra practice. Some of the problems are really easy, but others can be tricky.

http://math.mit.edu/~sswatson/pdfs/qualifying_round_2013.pdf

http://math.mit.edu/~sswatson/pdfs/qualifying_round_2014.pdf

Answer 30

oh......yay

Answer 31

I was thinking to finish with a last prob for today.

Answer 32

LOL I HAVE TRIED THAT PAPER...

Answer 33

best reference

Answer 34

@fbi2015 dont stop