What is the integral of 1/(t-1/2) ?

The integral of 1/(t-1/2) is ln|t-1/2 |+C

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integral of 1/(t-1/2)

\int \frac{1}{t - \frac{1}{2}} d t

ln t - \frac{1}{2} + C

Solution steps

\int \frac{1}{t - \frac{1}{2}} d t

Apply u-substitution

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = t - \frac{1}{2}

= ln t - \frac{1}{2}

Add a constant to the solution

= ln t - \frac{1}{2} + C

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