To apply the integral test to ∑n=2∞1n(lnn)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}∑n=2∞n(lnn)21, which integral should you evaluate?
∫2∞1n(lnn)2dn\int_2^{\infty} \frac{1}{n(\ln n)^2} dn∫2∞n(lnn)21dn
∫2∞1x(lnx)2dx\int_2^{\infty} \frac{1}{x(\ln x)^2} dx∫2∞x(lnx)21dx
∫0∞1n(lnn)2dn\int_0^{\infty} \frac{1}{n(\ln n)^2} dn∫0∞n(lnn)21dn
∫1∞1nlnndx\int_1^{\infty} \frac{1}{n \ln n} dx∫1∞nlnn1dx