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

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

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

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

2 arctan (x - 2) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = x - 2

= 2 arctan (x - 2)

Add a constant to the solution

= 2 arctan (x - 2) + C

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