integrate 1+xsinx+cosx/x (1+cosx)

I = (1 + xsinx + cosx )/ x (1+cosx)
= (1 + cosx + x sinx ) / x (1+cosx)

Individually dividing we get

=(1+cosx)/x (1+cosx) + ×sinx/x (1+cosx)
=1/x + sinx /(1+ cosx )

I1 = 1/x dx
= log|x| +c1

I2 = sinx/1+cosx dx

let 1+cosx = t
》 -sinx dx = dt
》sinx dx = -dt
I2 = -1/t dt
= - log |t| + c2
= - log|1+cosx| +c2

I = I1 + I2
= log|x| - log|1+cosx| + C
= log |[x/(1+cosx)]| + C