Find ∫(logx(x+1)²)dx

Applying integration by parts, we get:
∫logxdx/(x+1)² = logx∫dx/(x+1)² - ∫{1/x∫dx/(x+1)²}dx
= -logx/(x+1) + ∫dx/{x(x+1)}
= -logx/(x+1) + ∫(1/x - 1/(x+1))dx
= -logx/(x+1) + logx - log(x+1) + C