Question

Indefinite intergral question.

Answer 1

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

Answer 2

Um so for this would I just find the anti-derivative and then have that be the answer? (+C at the end)?

Answer 3

\[\int\limits x^{-3} dx+\int\limits x^3 dx \\ \text{ just use } \int\limits x^{n} dx=\frac{x^{n+1}}{n+1}+C \text{ for both }\]

Answer 4

\[\int\limits_{}^{}[(\frac{ x^{-2} }{ -2 })+(\frac{ x^4 }{ 4 })+C]\] would be the answer then?

Answer 5

drop integral sign and then I will say yes

Answer 6

Oh yes, pardon. Okay ^-^

Answer 7

you could write without negative exponent but that is about all you can do almost

Answer 8

\[- \frac{1}{2x^2}+\frac{x^4}{4}+C\]

Answer 9

ah so just further simplifying.

Answer 10

yeah not totally needed you can do other things but we don't have to get too crazy

Answer 11

\[\frac{1}{2x^2}(\frac{x^6}{2}-1)+C\]

Answer 12

\[\frac{1}{2x^2}(\frac{x^6-2}{2})+C \\ \frac{x^6-2}{4x^2}+C\]

Answer 13

but seriously you could just stop way earlier

Answer 14

haha yeah xD I don't think the prof expects too much from these.

Answer 15

Ok so for one with a little more stuff, like..
[intergral][(x-3)(2x+1)]dx

the answer would come out to
\[\frac{ 2x^3 }{ 3 }-\frac{ 5x^2 }{ 2 }-3x+C\] yeah?

Answer 16

*claps*

Answer 17

Omg. No joke. Freckles is the boss.

Answer 18

You're right

Answer 19

for both

Answer 20

the math too, you mean?

Answer 21

yes

Answer 22

yaya! thanks!

Answer 23

Working on other ones xD bleh fraction ones.