What is the integral of `int4z/(1+z^4)dx`

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What is the integral of `int4z/(1+z^4)dx`

Postby Integr88 » Wed 06 April, 2011 06:49

How to integrate `int4z/(1+z^4)dx`
it is the same if you sak What is the integral of `4z/(1+z^4)` or `4x/(1+x^4)` ?
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Re: What is the integral of `int4z/(1+z^4)dx`

Postby barnamah » Wed 06 April, 2011 08:09

to integrate `int4z/(1+z^4)dz`
now at the denominator we have `1+z^4` we know `z^4=(z^2)^2` so the denominator can be written as `1+(z^2)^2` so let's wire it in this format
`int4z/(1+(z^2)^2)dz`

let `u=z^2`
`du=2zdz` divide both side by 2 and we have:
`(du)/2=zdz`

now sub in the values and we got:
`int(4/((1+u^2))*1/2)du`

which we write as `int(1/((1+u^2))*4/2)du` and simplify to `int(1/((1+u^2))*2)du`
because 2 is constant, we can move it outside integral and we have:
`2int(1/(1+u^2))du`

Now it is simple. We know that `d/dxarctanx=1/(1+x^2)`
so if `1/(1+x^2)` is the derivative of `arctanx` we can use it here
`2int(1/(1+u^2))du=2(arctan(u))+C` where C is a constant.

we had `u=z^2` so sub it in and we have `2(arctan(z^2))+C`
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