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

The integral of 1/(t^2-t+2) is 2/(sqrt(7))arctan((2t-1)/(sqrt(7)))+C

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

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

\frac{2}{7} arctan (\frac{2 t - 1}{7}) + C

Solution steps

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

Complete the square

t^{2} - t + 2 : (t - \frac{1}{2})^{2} + \frac{7}{4}

= \int \frac{1}{( t - \frac{1}{2} ) ^{2} + \frac{7}{4}} d t

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 7} d u

= \int \frac{4}{4 u ^{2} + 7} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 7} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 7 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 7} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 4 \cdot \frac{1}{2 7} arctan (v)

Substitute back

= 4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (t - \frac{1}{2}))

Simplify

4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (t - \frac{1}{2})) : \frac{2}{7} arctan (\frac{2 t - 1}{7})

= \frac{2}{7} arctan (\frac{2 t - 1}{7})

Add a constant to the solution

= \frac{2}{7} arctan (\frac{2 t - 1}{7}) + C

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What is the integral of 1/(t^2-t+2) ?
The integral of 1/(t^2-t+2) is 2/(sqrt(7))arctan((2t-1)/(sqrt(7)))+C