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

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

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

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

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

Solution steps

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

Complete the square

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

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

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} (x - \frac{1}{2}))

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{x ^{2} ( x ^{2} - 1 )} d x

(t^{2} + 16)^{2} y^{^{'}} + 2 t (t^{2} + 16) y = 4, y (- 2) = 10

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

\int 36 - 4 x^{2} d x

\int 3 x e^{- 3 x} d x

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