The value of ∫lnn(1−(1x))dxx(x−1) is
Consider, I=∫ln(1−1x)x(x−1)dx
I=∫ln(1−1x)x2(1−1x)dx
Put ln(1−1x)=t ⇒ 1(1−1x)×1x2dx=dt
I=∫tdt
I=t22+c
I=12[ln(1−1x)]2+c
The solution set x= (b1c2−b2c1)(a1b2−a2b1) and y= (c1a2−c2a1)(a1b2−a2b1) to the pair of linear equations given below can be written as:
a1x + b1y + c1=0 and a2x + b2y + c2=0 .