What is the integral from 0 to P of 1/(M-P) ?

The integral from 0 to P of 1/(M-P) is -ln|M-P|+ln|M|

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integral from 0 to P of 1/(M-P)

\int_{0}^{P} \frac{1}{M - P} d P

- ln ∣ M - P ∣ + ln ∣ M ∣

Solution steps

\int_{0}^{P} \frac{1}{M - P} d P

Apply u-substitution

: \int_{M}^{M - P} - \frac{1}{u} d u

= \int_{M}^{M - P} - \frac{1}{u} d u

Take the constant out:

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

= - \int_{M}^{M - P} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{M}^{M - P}

Compute the boundaries:

ln ∣ M - P ∣ - ln ∣ M ∣

= - (ln ∣ M - P ∣ - ln ∣ M ∣)

Simplify

= - ln ∣ M - P ∣ + ln ∣ M ∣

(1) (0.3)

\int_{- 1}^{1} x^{3} (1 + x^{4})^{3} d x

f (x) = 8 + 5 x^{2} - 2 x^{3}, (1, 11)

f (x) = \frac{4}{x}, (4, 1)

\int_{0}^{l n (3)} 2 (3 - x e^{x}) d x

What is the integral from 0 to P of 1/(M-P) ?
The integral from 0 to P of 1/(M-P) is -ln|M-P|+ln|M|