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

The integral of 1/((-u^2+u-1)) is -2/(sqrt(3))arctan((2u-1)/(sqrt(3)))+C

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

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

- \frac{2}{3} arctan (\frac{2 u - 1}{3}) + C

Solution steps

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

Factor

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

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

Take the constant out:

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

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

Complete the square

u^{2} - u + 1 : (u - \frac{1}{2})^{2} + \frac{3}{4}

= - \int \frac{1}{( u - \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} + 3} d v

= - \int \frac{4}{4 v ^{2} + 3} d v

Take the constant out:

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

= - 4 \cdot \int \frac{1}{4 v ^{2} + 3} d v

Apply Integral Substitution

= - 4 \cdot \int \frac{1}{2 3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= - 4 \cdot \frac{1}{2 3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= - 4 \cdot \frac{1}{2 3} arctan (w)

Substitute back

= - 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (u - \frac{1}{2}))

Simplify

- 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (u - \frac{1}{2})) : - \frac{2}{3} arctan (\frac{2 u - 1}{3})

= - \frac{2}{3} arctan (\frac{2 u - 1}{3})

Add a constant to the solution

= - \frac{2}{3} arctan (\frac{2 u - 1}{3}) + C

\frac{d}{d x} (e^{c o s (x)} + cos (e^{x}))

(1 + cos^{2} (x))^{3}

(1 + x^{2}) \frac{d y}{d x} = 2 x y + 1

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

\frac{d}{d x} (\frac{16}{x ^{2}})

What is the integral of 1/((-u^2+u-1)) ?
The integral of 1/((-u^2+u-1)) is -2/(sqrt(3))arctan((2u-1)/(sqrt(3)))+C