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log(1+x^(2))"w.r.t."tan^(-1)x...

`log(1+x^(2))"w.r.t."tan^(-1)x`

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Let `u=log(1+x^(2))andv=tan^(-1)x`
Then we want to find `(du)/(dv)`
Differentiating u and v w.r.t. x, we get,
`(du)/(dx)=(d)/(dx)[log(1+x^(2))]=(1)/(1+x^(2)).(d)/(dx)(1+x^(2))`
`=(1)/(1+x^(2)).(0+2x)=(2x)/(1+x^(2))`
`(du)/(dx)=(d)/(dx)[log(1+x^(2))]=(1)/(1+x^(2)).(d)/(dx)(1+x^(2))`
`=(1)/(1+x^(2)).(0+2x)=(2x)/(1+x^(2))`
`therefore(du)/(dv)=((du//dx))/((dv//dx))=(((2x)/(1+x^(2))))/(((1)/(1+x^(2))))=2x`
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