`int_(1)^(2)(1)/(x(x^(2)+1))dx` का मान ज्ञात कीजिएः

Explanation

माना `(1)/(x(x^(2)+1))=A/x+(Bx+C)/(x^(2)+1)`
`rArr 1=A(x^(2)+1)+(Bx+C)x`
`rArr 1=Ax^(2)+Bx^(2)+Cx+A`
`rArr1=(A+B)x^(2)+Cx+A`
दोनों पक्षों से x के उचित घातको की तुलना करने पर,
A+B=0, C=0, A=1
`rArr B=-A, C=0, A=1`
`therefore 1/(x(x^(2)+1))=1/x+(-x)/(x^(2)+1)=1/x-(x)/(x^(2)+1)`
`rArr underset(1)overset(2)int (1)/(x(x^(2)+1))dx`
`=underset(1)overset(2)int [1/x-x/(x^(2)+1)]dx`
`=underset(1)overset(2)int 1/xdx-1/2underset(1)overset(2)int (2x)/(x^(2)+1)dx`
`=[log x]_(1)^(2)-1/2[log(x^(2)+1)]_(1)^(2)`
`=log2-1/2log5+1/2log2`
`=3/2 log2-1/2log5`