`int_(0)^(1) (2x+3)/(5x^(2)+1)` का मान ज्ञात कीजिएः

Explanation

`underset(0)overset(1)int (2x+3)/(5x^(2)+1)`
`=int(2x)/(5x^(2)+1)dx+3int(1)/(5x^(2)+1)dx`
`=int(2x)/(5x^(2)+1)dx+3int(1)/(x^2+(1/5))dx`
माना `5x^(2)+1=t Rightarrow 10xdx =dt "या "2xdx=1/5dt`
`therefore (2x+3)/(5x^2+1)dx`
`=1/5int (dt)/(t)+3/5int(1)/(x^(2)+((1)/(sqrt5))^(2))dx`
`=1/5log|5x^(2)+1|+3/sqrt5tan^(-1)(sqrt5x)`
`rArr underset(0)overset(1)int (2x+3)/(5x^(2)+1)dx`
`=[1/5log(5x^(2)+1)+3/sqrt5tan6(-1)(sqrt5x)]_(0)^(1)`
`=[{1/5log(5xx1^(2)+1)+(3)/sqrt5tan^(-1)(sqrt5xx1)}-{1/5log(5xx0+1)+3/sqrt5tan^(-1)(sqrt5xx0}]`
`=1/5log6+3/sqrt5tan^(-1)(sqrt5) -{1/5log 1+(3)/sqrt5tan^(-1)(0)}`
`=1/5log 6+3/sqrt5tan^(-1)(sqrt5)-(0+0)`
`=1/5log6+3/sqrt5tan^(-1)(sqrt5)`