If ∫1−x7x1+x7dx=aln⁡|x|+bln⁡x7+1+c, then

I=∫1−x7x1+x7dx=aln⁡|x|+bln⁡1+x7+C Differentiating both sides, we get 1−x7x1+x7=ax+b⋅7x61+x7⇒ 1−x7=a1+x7+7bx7⇒ a=1,a+7b=−1⇒ b=−27