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inte^(tan^-1x)(1+x/(1+x^2))dx...

`inte^(tan^-1x)(1+x/(1+x^2))dx`

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Let `I = inte^(tan^(-1)x) ((1+x+x^(2))/(1+x^(2)))dx`
`=inte^(tan^(-1)x)((1+x^(2))/(1+x^(2))+(x)/(1+x^(2)))dx`
`= inte^(tan^(-1)).^(x)dx + int (xe^(tan^(-1)x))/(1+x^(2))dx`
`I = I_(1) + I_(2)`
Now, `I_(2) = int (xe^(tan^(-1)x))/(1+x^(2)) =dx`
Put `tan^(-x) = t rArr x = tant `
` rArr 1/(1+x^(2)) dx`
`rArr 1/(1+x^(2)) dx = dt`
` :. I = int underset(I)tant.underset(II)(e^(t)dt)`
`= tan t.e^(t)-int sec^(2)t.e^(t)dt+C`
` =tant.e^(t) - int(1+tan^(2)t) e^(t) dt + C [:' sec^(2) theta = 1 + tan^(2)theta]`
`I_(2)= tant.e^(t)-int(1+x^(2))(e^(tan^(-1)x))/(1+x^(2))dx + C`
`I_(2) = tant. e^(t) -inte^(tan^(-t))dx + C`
`:. I = inte^(tan^(-1)).^(x)dx + tant.e^(t) - inte^(tan^(-1)).^(x)dx + C`
`= tant.e^(t) + C`
`= xe^(tan^(-1)x)+ C`
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