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`int(4x-3)/(3x^(2)+2x-5)` का मान ज्ञात कीजिए ।

Answer» माना `" "4x-3=A(d)/(dx)(3x^(2)+2x-5)+B`
`rArr" "4x-3=A(6x+2)+B`
दोनों और x के समान पदों के गुणकों की तुलना करने पर
`A=(2)/(3),B=-(13)/(3)`
`therefore" "int(4x-3)/(3x^(2)+2x-5)dx=int((2)/(3)(6x+2)+(-(13)/(3)))/(3x^(2)+2x-5)dx`
`=(2)/(3)log(3x^(2)+2x-5)-(13)/(3).(1)/(3)int(1)/(x^(2)+(2)/(3)x-(5)/(3))dx`
`=(2)/(3)log(3x^(2)+2x-5)-(13)/(9)int(1)/((x+(1)/(3))^(2)-((4)/(3))^(2))dx`
`=(2)/(3)log (3x^(2)+2x-5)-(13)/(9).(1)/(2.(4)/(3))log[((x+(1)/(3))-(4)/(3))/((x+(1)/(3))+(4)/(3))]+c`
`=(2)/(3)log(3x^(2)+2x-5)_(13)/(24)log((3x-3)/(3x+5))+c`


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