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

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

Popular >

integral from 0 to a of 1/((a-x))

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

diverges

Solution steps

\int_{0}^{a} \frac{1}{a - x} d x

Apply u-substitution

: \int_{a}^{0} - \frac{1}{u} d u

= \int_{a}^{0} - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{a}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{a}^{0}

Compute the boundaries:

diverges

= diverges

\int_{0}^{5} (2 e^{0.2 x} - x) d x

\int_{0}^{l n (11)} tanh (x) d x

\int_{0}^{1} \frac{( 2 x ^{2} - 5 )}{x} d x

\int_{0}^{1.2} e^{- x} cos (x) d x

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

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