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

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

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integral from-1 to 2 of (x+1)/(x(x+1)^2)

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 0

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

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

\int_{2 y}^{1 + y^{2}} z d z

\int_{- 2 π}^{2 π} 9 sin (x) d x

\int_{0}^{x^{2}} y d y

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