What is the integral of 1/(2x-x^2-10) ?

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

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integral of 1/(2x-x^2-10)

\int \frac{1}{2 x - x ^{2} - 10} d x

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

Solution steps

\int \frac{1}{2 x - x ^{2} - 10} d x

Complete the square

2 x - x^{2} - 10 : - (x - 1)^{2} - 9

= \int \frac{1}{- ( x - 1 ) ^{2} - 9} d x

Apply u-substitution

= \int \frac{1}{- u ^{2} - 9} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{3} arctan (v)

Substitute back

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

Add a constant to the solution

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

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