`(2x)/((x^(2)+1)(x^(2)+3))`

Explanation

माना `I=int(2x)/(x^(2)+1)(x^(2)+3))dx`
माना `x^(2)=t" "rArr" "2xdx=dt`
`therefore" "I=int(1dt)/((t+1)(t+3))=int(1)/((t+1)(t+3))dt`
माना `(1)/((t+1)(t+3))=(A)/(t+1)+(B)/(t+3)`
`rArr" "(1)/((t+1)(t+3))=(A(t+3)+B(t+1))/((t+1)(t+3))`
`rArr" "1=A(t+3)+B(t+1)`
`{:(t=-1,"तो ",1=A(-1+3)+0,rArr.,A=(1)/(2)),(t=-3,"तो ",1=0+B(-3+1),rArr.,B=(1)/(2)):}`
`therefore" "I=(1)/(2)int(1)/(t+1)dt-(1)/(2)int(1)/(t+3)dt`
`" "=(1)/(2)log|t+1|-(1)/(2)log|t+3|+C`
`" "=(1)/(2)log|(t+1)/(t+3)|+C=(1)/(2)log|(x^(2)+1)/(x^(2)+3)|+C`