What is the integral from 1 to infinity of (x^2+x+1) ?

The integral from 1 to infinity of (x^2+x+1) is diverges

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integral from 1 to infinity of (x^2+x+1)

\int_{1}^{\infty} (x^{2} + x + 1) d x

diverges

Solution steps

\int_{1}^{\infty} x^{2} + x + 1 d x

Compute the indefinite integral:

\int x^{2} + x + 1 d x = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + C

Compute the boundaries:

\int_{1}^{\infty} (x^{2} + x + 1) d x =

diverges

= diverges

\int_{0}^{1} \frac{x}{( x ^{2} + 3 ) ^{3}} d x

\int_{0}^{s i n (2 x)} arcsin (t) d t

\int_{2}^{3} x^{2} + x + 1 d x

\int_{x}^{\infty} e^{- x} d x

\int_{1}^{3} x^{2} - 6 x + 5 d x

What is the integral from 1 to infinity of (x^2+x+1) ?
The integral from 1 to infinity of (x^2+x+1) is diverges