Question

how can i start this integral?

Answer 1

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

Answer 2

simplify the denominator by multiplying.

Answer 3

Or partial fractions might be better . . .

Answer 4

how would i do that?

Answer 5

This one is a little tricky.
You have a repeated quadratic factor.\[\Large\bf\sf \frac{1}{(x^2+1)^2}\quad=\quad \frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}\]

Answer 6

Trig sub seems better for this problem actually D:
But I dunno..

Answer 7

tried x = tan t ?

Answer 8

Have you learned about Trig sub yet Big Money? :o
Works out really nicely on this one.

Answer 9

yea sort of

Answer 10

if x=tan t dx=sec^2 x t

Answer 11

sec^2 x t ??
How did that x get in there? D:

Answer 12

that is what hartnn told me to try

Answer 13

my bad is sec^2 t?

Answer 14

ya that looks good.
dx = sec^2 t

Answer 15

\[\Large\bf\sf \int\limits \frac{1}{(1+\color{orangered}{x}^2)^2}\;\color{orangered}{dx}\]Plug your stuff in dude +_+

Answer 16

better you convert everything into sin and cos in your new integral

Answer 17

is it\[\int\limits_{0}^{1} \frac{ \sec ^{2} t}{ (\tan ^{2} t+1 )^{2}}\]

Answer 18

Couple things to fix real quick.\[\Large\bf\sf \int\limits\limits_{\color{red}{\cancel{0}}}^{\color{red}{\cancel{1}}} \frac{ \sec ^{2} t}{ (\tan ^{2} t+1 )^{2}}\color{red}{dt}\]

Answer 19

Those boundaries are for x, we can't use them for t.

Answer 20

Understand how to simplify the integrand?
Use your square identity! :O\[\Large\bf\sf \color{royalblue}{\tan^2t+1\quad=\quad \sec^2t}\]

Answer 21

would it be \[\int\limits \frac{ \sec ^{2}t }{ (\sec ^{2}t)^{2} }dt\]

Answer 22

yes, go ahead
find the new limits too

Answer 23

would the limits be from 0 to pi/2

Answer 24

x = tan t

t = arctan x
when x = 0
t = 0

when x = 1
t = arctan 1 = pi/4

Answer 25

do i plug them in already, or do i need to do something else?

Answer 26

\(\int \limits_0^{\pi/4}\dfrac{1}{\sec^2t}dt\)
\(=\int \limits_0^{\pi/4}\cos^2tdt\)
can you integrate that ?