Question

What is the integral of `x/(1+x^2)`?

Answer 1

`intx/(x^2+1)dx=1/2ln(x^2+1)+C`

Explanation:

Let `u(x)=1+x^2" "` then `" "du(x) =2xdx`
`" "`
`color(blue)((d(u(x)))/2=xdx)`
`" "`
Start solving the integral.
`" "`
`intx/(x^2+1)dx`
`" "`
`=intcolor(blue)((d(u(x)))/(2u(x))`
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`=1/2int(du(x))/(u(x))`
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`=1/2lnabs(u(x))+C`
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`=1/2lnabs(x^2+1)+C`
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Because `x^2+1>0 " " then " " abs(x^2+1)=x^2+1`
`" "`
Therefore,
`" "`
`intx/(x^2+1)dx=1/2ln(x^2+1)+C`