What is the integral from 0 to 6 of 1/((4-x)) ?

The integral from 0 to 6 of 1/((4-x)) is diverges

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integral from 0 to 6 of 1/((4-x))

\int_{0}^{6} \frac{1}{( 4 - x )} d x

diverges

Solution steps

\int_{0}^{6} \frac{1}{4 - x} d x

Find undefined (singularity) points:

x = 4

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_{0}^{4} \frac{1}{4 - x} d x + \int_{4}^{6} \frac{1}{4 - x} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{2} 2 t ln (1 + t) d t

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

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

\int_{0}^{1} 2 x e^{- 2 x} d x

\int_{0}^{1} - 2 d x

What is the integral from 0 to 6 of 1/((4-x)) ?
The integral from 0 to 6 of 1/((4-x)) is diverges