What is the integral of 1/(-x^2+4x-5) ?

The integral of 1/(-x^2+4x-5) is -arctan(x-2)+C

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integral of 1/(-x^2+4x-5)

\int \frac{1}{- x ^{2} + 4 x - 5} d x

- arctan (x - 2) + C

Solution steps

\int \frac{1}{- x ^{2} + 4 x - 5} d x

Complete the square

- x^{2} + 4 x - 5 : - (x - 2)^{2} - 1

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

Apply u-substitution

= \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 - 2

= - arctan (x - 2)

Add a constant to the solution

= - arctan (x - 2) + C

What is the integral of 1/(-x^2+4x-5) ?
The integral of 1/(-x^2+4x-5) is -arctan(x-2)+C