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Evaluate intcot^(-1)(x^(2)+x+1)dx...

Evaluate
`intcot^(-1)(x^(2)+x+1)dx`

Text Solution

Verified by Experts

We have
`intcot^(-1)(x^(2)+x+1)dx=inttan^(-1)((1)/(x^(2)+x+1))dx`
`=inttan^(-1)((x+1-x)/(1+(x+1)x))dx`
`=inttan^(-1)(x+1)dx-inttan^(-1)xdx`
`=xtan^(-1)(x+1)-intx*(1)/(1+(x+1)^(2))dx-{xtan^(-1)x-int(x)/(1+x^(2))dx}`
`=x{tan^(-1)(x+1)-tan^(-1)x}+int(x)/(1+x^(2))dx-int(x)/(1+(1+x))^(2)dx`
`=x{tan^(-1)(x+1)-tan^(-1)x}+(1)/(2)ln(1+x^(2))-int(x)/(1+(1+x)^(2))dx`
Now `int(x)/(1+(1+x)^(2))dx=int(z-1)/(1+z^(2))dz`, `z=1+x`
`=int(z)/(1+z^(2))dz-int(1)/(1+z^(2))dz`
`=(1)/(2)ln|1+z^(2)|-tan^(-1)(z)`
`=(1)/(2)ln|1+(1+x)^(2)|-tan^(-1)(1+x)`
Hence ,
`intcot^(-1)(x^(2)+x+1)dx=x{tan^(-1)(x+1)-tan^(-1)x}+(1)/(2)log(1+x^(2))-(1)/(2)log(1+(1+x^(2))+tan^(-1)(1+x)+C`
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