Question

Find the anti-derivative of

Answer 1

\[f(x) = x/(x ^{2} + 1)^{2}\]

Answer 2

that appears to be an arctan

Answer 3

x^2 + 1 = 2x we need a 2 up top to ln(x) it

Answer 4

multiply by 1; or rather 2/2 keep the top but pull out the 1/2

Answer 5

that might not be good tho :)

Answer 6

x
----------
(x^2 +1)^2


tan the bottom :) maybe

Answer 7

*alternate method time*
1/2 * f 2x/(x^4+1)
u=x^2
du=2x
a=1

arctan(x^2)+C

Answer 8

tan(t) = x

tan^2(t) = x^2

tan^2 + 1 = sec^2

sec^2^2 = sec^4

tan(t) sec^4(t)

Answer 9

here is the rule I used:

Answer 10

tan(t) sec^-4(t) maybe :)

Answer 11

my way you would still have to convert dx to dt...

Answer 12

\[\int\limits_{ }^{ }1/ (a^2+u^2) = 1/a \arctan(u/a)+C\]

Answer 13

well, that equation looks terrible...

Answer 14

wow lol

Answer 15

tan(t) = x

dt sec^2 = dx thats helpful to me i think

Answer 16

[S] tan(t) sec^-2(t) dt ....maybe

Answer 17

tan = sin/cos = sin sec-1..... mines just messy lol

Answer 18

Isn't that straight u substitution? f(x) = x (x^2 + 1) ^ -2
Let u= x^2 + 1 and du = 2x

Answer 19

[S] 1/2u^2 du ?

Answer 20

That would give you...\[^{} \int\limits_{}^{} 1/2u ^{2}du\]

Answer 21

u^-2 = -u^-1

Answer 22

I mean -2 as the expontent

Answer 23

-1/2u = -1/2(x^2+1) +C

Answer 24

\[1/2\int\limits_{}^{}u ^{-2}du\]

Answer 25

=-1/(2u) + C, then substitue x^2 + 1 back in for u