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

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

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

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

ln x^{2} - 4 x + 5 + 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 = ln u^{2} + 1 + u

= ln u^{2} + 1 + u

Substitute back

u = x - 2

= ln (x - 2)^{2} + 1 + x - 2

Expand

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

= ln x^{2} - 4 x + 5 + x - 2

Add a constant to the solution

= ln x^{2} - 4 x + 5 + x - 2 + C

x^{2} y^{^{'}} + 2 x y = 15 cos^{2} (x)

\int_{- 1}^{2} x^{2} - 2 x d x

f (x) = 4 - x

x \to 0 + lim (x^{3 s i n (x)})

\int (\frac{1}{4})^{x + 1} d x

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