limn→∞∑r=1n(1+n2r2)(1+nr)n1=∫01(1+x2)(1+x)dx
Let I=∫01(1+x2)(1+x)dx
Now x=tanθ,dx=sec2\thetadθ
I=∫04π1+tanθdθ=21∫04πsinθ+cosθ2cos\thetadθ
=21∫04πsinθ+cosθ(sinθ+cosθ)dθ+21∫04πsinθ+cosθ(cosθ−sinθ)dθ
=21×4π+21[ln∣sinθ+cosθ∣]04π =8π+41ln2