The value of ∫018log(1+x)1+x2dx is
π8log2
π2log2
log2
πlog2
The explanation for the correct option.
Evaluate the integral.
∫018log(1+x)1+x2dxx=tanθsec2θdθ=dxθ=0,x=0,θ=π4I=8∫0π4log(1+tanθ)1+tan2θsec2θdθ=8∫0π4log(1+tanθ)dθ=8∫0π4log21+tanθdθ=8∫0π4log2-log(1+tanθ)dθ=8log2×π4=πlog2
Hence, option D is correct .
The value of is
A. 1
B. 0
C. − 1
D.