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

- \frac{1}{( 3 + t ) ^{2}}

t^{2} e^{-, at}

\int \frac{x ^{2} + 25}{x} d x

\int (7 - \frac{1}{3} sin (x) + 9 cos (x)) d x

f (x) = x + \frac{17}{x}, at x = 1

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