`int_(0)^(a) (dx)/(ax+a^(2)-x^(2))` का मान ज्ञात कीजिएः।

Explanation

`underset(0)overset(a)int (dx)/(ax+a^(2)-x^(2))`
`=underset(0)overset(a)int (dx)/(x^(2)-ax-a^(2))`
`=underset(0)overset(a)int (dx)/(x^(2)-ax+((a)/(2))^(2)-((a)/(2))^(2)-a^(2))`
`=underset(0)overset(a)int (dx)/((x-a/2)^(2)-(a^(2))/(4)-a^2)`
`=-underset(0)overset(a)int (dx)/((x-a/2)^(2)-(5a^(2))/4)`
`=-underset(0)overset(a)int (dx)/((x-a/2)^(2)-((sqrt5a)/(2))^(2))`
`=-1/(2xx((sqrt5a)/(2)))[log|((x-a/2)-((sqrt5a)/(2)))/((x-a/2)+((sqrt5a)/(2)))|]_(0)^(a)`
`=-1/(sqrt5a)[log|(2x-a-sqrt5a)/(2x-a+sqrt5a)|]_(0)^(a)`
`=-1/(sqrt5a)[log|(2x-a-sqrt5a)/(2x-a+sqrt5a)|-log|(0-a-sqrt5a)/(0-a+sqrt5a)|]`
`=-1/(sqrt5a)[log|(a-sqrt5a)/(a+sqrt5a)|-log|(a+sqrt5a)/(a-sqrt5a)|]`
`=-1/(sqrt5a)[log|(1-sqrt5)/(1+sqrt5)|-log|(1+sqrt5)/(1-sqrt5)|]`
`=-1/(sqrt5a)log|(1-sqrt5)/(1+sqrt5)xx(1-sqrt5)/(1+sqrt5)|`
`=-1/(sqrt5a)log|((1-sqrt5)^(2))/((1+sqrt5))|`
`=-1/(sqrt5a)log|(1+5+2sqrt5)/(1+5-2sqrt5)|`
`=-1/(sqrt5a)log|(6+2sqrt5)/(6-2sqrt5)|`
`=-1/(sqrt5a)log|(3+sqrt5)/(3-sqrt5)|`