What is the integral from-1 to 1 of 4/x ?

The integral from-1 to 1 of 4/x is diverges

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integral from-1 to 1 of 4/x

\int_{- 1}^{1} \frac{4}{x} d x

diverges

Solution steps

\int_{- 1}^{1} \frac{4}{x} 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{4}{x} d x + \int_{0}^{1} \frac{4}{x} d x

\int_{- 1}^{0} \frac{4}{x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{\partial}{\partial x} (x^{\frac{1}{8}} \cdot y^{\frac{11}{9}})

\int \frac{x ^{5}}{( 1 - x ^{3} ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (1 - (x^{2} + y^{2}))

(x - y) d x - x d y = 0

\int x^{3} x^{2} + 3 d x

What is the integral from-1 to 1 of 4/x ?
The integral from-1 to 1 of 4/x is diverges