Anish Singhal
Last Activity: 6 Years ago
Given ∫{log(logx) + 1/(logx)2}dx
we know that ∫ U.V dx = U ∫ V dx - ∫ ( d( U)/dx ∫ V dx ) dx.
⇒ ∫{log(logx) × 1 dx + ∫1/(logx)2dx
⇒ log(logx) ∫ 1 dx - ∫ ( d( log(logx))/dx ∫ 1 dx ) dx + ∫1/(logx)2dx [ d( log(logx))/dx = 1 / x × log x ]
⇒ log(logx) × x - ∫ [1 /x× logx] ×x) dx + ∫1/(logx)2dx
⇒ log(logx) × x - ∫ (1 / logx)× 1 dx + ∫1/(logx)2dx
⇒ log(logx) × x - [ (1 / logx)× ∫1 dx -∫d(1/log x)/ dx ∫ 1. dx] + ∫1/(logx)2dx
⇒ log(logx) × x - [ (1 / logx)× x -∫ [-1/(log x)2× (1/x) ×x] + ∫1/(logx)2dx
⇒ log(logx) × x - [ (1 / logx)× x + ∫ [1/(log x)2] + ∫1/(logx)2dx
⇒ log(logx) × x - (1 / logx)× x-∫ [1/(log x)2]+∫1/(logx)2dx
∴ x log(log x) - x / log x.