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

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

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integral of 1/(2t-t^2)

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

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

Solution steps

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

Complete the square

2 t - t^{2} : - (t - 1)^{2} + 1

= \int \frac{1}{- ( t - 1 ) ^{2} + 1} d t

Apply u-substitution

: \int \frac{1}{- u ^{2} + 1} d u

= \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

Substitute back

u = t - 1

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

Simplify

\frac{ln ∣ t - 1 + 1 ∣}{2} - \frac{ln ∣ t - 1 - 1 ∣}{2} : \frac{1}{2} ln ∣ t ∣ - \frac{1}{2} ln ∣ t - 2 ∣

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

Add a constant to the solution

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

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What is the integral of 1/(2t-t^2) ?
The integral of 1/(2t-t^2) is 1/2 ln|t|-1/2 ln|t-2|+C