`int 1/ sqrt (a^2 + x^2) dx = log ( x + sqrt(x^2 + a^2) + c`

Explanation

L.H.S. = ` I = int (1)/(sqrt(x^(2)) + a^(2))) dx `
Let ` x = a tan theta rArr tan theta = (x)/(a)`
` therefore dx = a sec^(2) theta . d theta `
`therefore I = int (1)/(sqrt(a^(2) . tan^(2) theta + a^(2))). a sec^(2) theta * d theta `
` = int (a*sec^(2) theta )/(a sqrt( 1 + tan^(2) theta)) d theta `
`= int (sec^(2) theta )/(sec theta) d theta = int sec theta * d theta `
` = log [ tan theta + sec theta ] + C_(1)`
` log [ (x)/(a) + sqrt(sec^(2) theta)] + C_(1)`
`[ " " because x = a tan theta ] `
` log [ (x)/(a) + sqrt( 1 + tan^(2) theta ] + C_(1)`
` = log [ (x)/(a) + sqrt(1 + (x^(2))/(a^(2)) ] + C_(1)`
` = log [ (x)/(a) + (sqrt(a^(2) + x^(2)))/(a) ] + C_(1)`
` log[x + sqrt(a^(2) + x^(2))] - log a + C_(1)`
`therefore int (1)/(sqrt(x^(2) + a^(2))) dx = log [ x + sqrt(x^(2) + a^(2))] + C = RHS `
where ` C = - log a + c_(1)` Hnece Proved