ज्ञात करें `int (sqrt(16+(log x)^2))/x ""dx `

Explanation

माना की `I =int (sqrt(16+(log x)^2))/x ""dx `
`log x = t ` रखे, तो `1/x ""dx =dt`
अब, `I = int sqrt(4^2+t^2) dt = ( t sqrt(4^2+t^2))/2 +4^2/2 "log"|t+sqrt(4^2 +t^2|+c`
` =(log x sqrt(16+(logx)^2))/2+8log|logx +sqrt(16+(logx)^2)+c`