What is the integral of 1/(-x^2+6x-10) ?

The integral of 1/(-x^2+6x-10) is -arctan(x-3)+C

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integral of 1/(-x^2+6x-10)

\int \frac{1}{- x ^{2} + 6 x - 10} d x

- arctan (x - 3) + C

Solution steps

\int \frac{1}{- x ^{2} + 6 x - 10} d x

Complete the square

- x^{2} + 6 x - 10 : - (x - 3)^{2} - 1

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

Apply u-substitution

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

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

Factor

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

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

Use the common integral:

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

= - arctan (u)

Substitute back

u = x - 3

= - arctan (x - 3)

Add a constant to the solution

= - arctan (x - 3) + C

\frac{d}{d x} ((sinh (x))^{2 x^{2} + 2 x})

\int \frac{( ln ( x ) ) ^{2}}{2} d x

y^{^{'}} = 6 y^{2} + x y^{2}, y (0) = 1

\frac{d ^{2}}{d x ^{2}} (\frac{2000 t}{4 t + 75})

\int (y cos (x) + 2 x e^{y}) d x

What is the integral of 1/(-x^2+6x-10) ?
The integral of 1/(-x^2+6x-10) is -arctan(x-3)+C