What is the integral from-2 to 0 of 1/((x+1)^2) ?

The integral from-2 to 0 of 1/((x+1)^2) is diverges

Popular >

integral from-2 to 0 of 1/((x+1)^2)

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = - 1

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{c o s (x)} x d x

\int_{0}^{1} x arcsin (x) d x

\int_{1}^{5} \frac{x - 1}{x} d x

\int_{\frac{π}{2}}^{π} sin (x) cos (x) d x

\int_{0}^{\frac{π}{4}} sin (x) cos (x) d x

What is the integral from-2 to 0 of 1/((x+1)^2) ?
The integral from-2 to 0 of 1/((x+1)^2) is diverges