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Evaluate: int1/(1+x+x^2+x^3)dx...

Evaluate: `int1/(1+x+x^2+x^3)dx`

Text Solution

Verified by Experts

Given that,
`I=int1/(1+x+x^2+x^3)dx`
`I=int1/((x+1)(1+x^2))`
so,`1/((x+1)(1+x^2))`
`=A/(x+1)+(Bx+C)/(x^2+1)`
`=>1=A(x^2+1)+(Bx+c)(x+1)`
if `x=-1` then
`1=2A+0`
`=>A=1/2`
if `x=0` then
`1=A+C`
`=>C=1/2`
`x=0` then
`1=2A+2B+2C`
`=>B=-1/2`
putting these we get,
`int 1/((x+1)(1+x^2))=int((1/2)/(x+1)+(1/2x)/(x^2+1)+1/2*1/(x^2+1))dx`
`=1/2 log (x+1) -1/4 log(x^2+1) +1/2 tan^-1 (x)+c`
`=1/2{log(x+1)/sqrt(x^2+1) +tan^-1 (x)+c}`
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