माना `I=int(dx)/(a^(2)-x^(2))=int(dx)/((a-x)(a+x))`
आंशिक भिन्नों में वियोजन `(1)/((a-x)(a+x))=(A)/((a-x))+(B)/((a+x))`
`rArr” “1=A(a+x)+B(a-x)” …(1)”`
माना समीकरण (1 ) के दोनों पक्षों में `a-x=0 rArr x = a` रखने पर,,
`1=A(2a)+0 rArr A=(1)/(2a)`
पुनः समीकरण (2 ) के दोनों पक्षों में `a+x=0 rArr x = -a`
रखने पर,
`1=0 +B(2a)” “therefore” “B=(1)/(2a)`
अतः `(1)/((a-x)(a+x))=(1)/(2a(a-x))+(1)/(2a(a+x))`
`therefore” “int(dx)/((a-x)(a+x))=(1)/(2a)int(dx)/((a-x))+(1)/(2a)int(dx)/((a+x))`
`=-(1)/(2a)log(a-x)+(1)/(2a)log(a+x)`
`=(1)/(2a)log((a+x)/(a-x))`
`(because log M-log N=log(M)/(N))`
`therefore” “int(dx)/(a^(2)-x^(2))=(1)/(2a)log((a+x)/(a-x)),` यदि `a gt x`

माना `I=int(dx)/(a^(2)+x^(2))=(1)/(a^(2))int(dx)/(1+((x)/(a))^(2))`
माना `(x)/(a)=tan theta`
`therefore” “dx=a sec^(2) theta d theta`
`therefore” “I=(a)/(a^(2))int(sec^(2) theta d theta)/(1+tan^(2) theta)=(1)/(a) int d theta =(1)/(a) theta`
अतः `int(dx)/(a^(2)+x^(2))=(1)/(a) tan^(-1)((x)/(a))`
`” “[because tan theta =(x)/(a)rArr theta = tan^(-1)((x)/(a))]`