What is the integral from 0 to 1 of 1/(mx+1) ?

The integral from 0 to 1 of 1/(mx+1) is (ln|m+1|)/m

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integral from 0 to 1 of 1/(mx+1)

\int_{0}^{1} \frac{1}{m x + 1} d x

\frac{ln ∣ m + 1 ∣}{m}

Solution steps

\int_{0}^{1} \frac{1}{m x + 1} d x

Apply u-substitution

: \int_{1}^{m + 1} \frac{1}{m u} d u

= \int_{1}^{m + 1} \frac{1}{m u} d u

Take the constant out:

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

= \frac{1}{m} \cdot \int_{1}^{m + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{m} [ln ∣ u ∣]_{1}^{m + 1}

Compute the boundaries:

ln ∣ m + 1 ∣

= \frac{1}{m} ln ∣ m + 1 ∣

Simplify

= \frac{ln ∣ m + 1 ∣}{m}

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What is the integral from 0 to 1 of 1/(mx+1) ?
The integral from 0 to 1 of 1/(mx+1) is (ln|m+1|)/m