What is the integral of 1/(-2+3x-5x^2) ?

The integral of 1/(-2+3x-5x^2) is -2/(sqrt(31))arctan((10x-3)/(sqrt(31)))+C

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integral of 1/(-2+3x-5x^2)

\int \frac{1}{- 2 + 3 x - 5 x ^{2}} d x

- \frac{2}{31} arctan (\frac{10 x - 3}{31}) + C

Solution steps

\int \frac{1}{- 2 + 3 x - 5 x ^{2}} d x

Complete the square

- 2 + 3 x - 5 x^{2} : - 5 (x - \frac{3}{10})^{2} - \frac{31}{20}

= \int \frac{1}{- 5 ( x - \frac{3}{10} ) ^{2} - \frac{31}{20}} d x

Apply u-substitution

: \int \frac{20}{- 100 u ^{2} - 31} d u

= \int \frac{20}{- 100 u ^{2} - 31} d u

Take the constant out:

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

= 20 \cdot \int \frac{1}{- 100 u ^{2} - 31} d u

Apply Integral Substitution

= 20 \cdot \int - \frac{1}{10 31 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 20 (- \frac{1}{10 31} \cdot \int \frac{1}{v ^{2} + 1} d v)

Use the common integral:

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

= 20 (- \frac{1}{10 31} arctan (v))

Substitute back

= 20 (- \frac{1}{10 31} arctan (\frac{10}{31} (x - \frac{3}{10})))

Simplify

20 (- \frac{1}{10 31} arctan (\frac{10}{31} (x - \frac{3}{10}))) : - \frac{2}{31} arctan (\frac{10 x - 3}{31})

= - \frac{2}{31} arctan (\frac{10 x - 3}{31})

Add a constant to the solution

= - \frac{2}{31} arctan (\frac{10 x - 3}{31}) + C

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What is the integral of 1/(-2+3x-5x^2) ?
The integral of 1/(-2+3x-5x^2) is -2/(sqrt(31))arctan((10x-3)/(sqrt(31)))+C