Question

integration of[1/(1+x^2)^2]dx​

Answer 1

Answer:

$\dfrac{ x}{2(1 + {x}^{2}) } + \dfrac{1}{2} \tan ^{- 1} x + c$

Step-by-step explanation:

Let x = tana

then

$dx = ({ \sec}^{2} a)da$

Now,

$\dfrac{1}{({1 + {x}^{2} )}^{2} } = \dfrac{1}{({1 + { \tan }^{2}a )}^{2} }$

which makes,

$\int \: \dfrac{1}{ { ( { \sec }^{2}a )}^{2} } ({ \sec }^{2}a) \: da \\ = \int ({ \cos}^{2}a) da$

Cos a can be expressed as: (cos 2a +1)/2

$\int (\dfrac{\cos(2a) + 1}{2} )da \\ = \int \dfrac{ \cos(2a) }{2} da + \int \dfrac{1}{2} da \\ = \dfrac{ \sin(2a) }{4} + \dfrac{1}{2} a + c$

Substituting for x = tana here,

$\dfrac{ \sin(2\tan ^{- 1} x) }{4} + \dfrac{1}{2} \tan ^{- 1} x + c$

Using

$2\tan ^{- 1} x = \tan ^{- 1} \dfrac{2x}{1 - {x}^{2} }$

And using the pythogorean triangle for tana here we get

$2\tan ^{- 1} x = \sin ^{- 1} ( \dfrac{2x}{1 + {x}^{2} } )$

Using this in above equation we get the anti derivative as:

$\dfrac{ x}{2(1 + {x}^{2}) } + \dfrac{1}{2} \tan ^{- 1} x + c$