`int (x tan^(-1) x)/((1+x^(2))^(3//2)) dx` का मान ज्ञात कीजिए।

Answer

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Explanation

दिया है : `int (x tan^(-1) x)/((1+x^(2))^(3//2)) dx` माना की `x=tan theta`, तब `dx=sec^(2) theta d theta` या `theta=tan^(-1) x`.
`int (theta. tan theta.sec^(2) theta d theta)/((1+tan^(2) theta)^(3//2))=int (theta.tan theta.sec^(2) theta d theta)/((sec^(2) theta)^(3//2))`
`=int (theta.tan theta.sec^(2) theta d theta)/(sec^(3) theta)`
`=int (theta.tan theta)/(sec theta) d theta`
`=int theta (sin theta)/(cos theta) .cos theta d theta`
`=int theta.sin theta d theta`
`=theta.int sin theta d theta-int {d/(d theta) theta.int sin theta d theta}d theta`
`=-theta cos theta+int cos theta d theta`
`=-theta cos theta +sin theta+c`
`=-tan^(-1) x.(1)/sqrt(1+x^(2))+(x)/sqrt(1+x^(2))+c`
`=(1)/sqrt(1+x^(2)) [-tan^(-1) x+x]+c`.