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

= \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

\int 8 x - x^{2} d x

\int x 12 - x^{2} d x

y = 1 - x^{2}, y = 4 - 4 x

f (x) = - 3 x

\frac{\partial}{\partial x} ((x^{2} y^{2})^{\frac{1}{3}})

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