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

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

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

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

ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5) + C

Solution steps

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

Complete the square

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

= \int \frac{1}{( x - 1 ) ^{2} + 4} d x

Apply u-substitution

= \int \frac{1}{u ^{2} + 4} d u

Apply Trigonometric Substitution

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{2} (x - 1))) + sec (arctan (\frac{1}{2} (x - 1)))

Simplify

ln tan (arctan (\frac{1}{2} (x - 1))) + sec (arctan (\frac{1}{2} (x - 1))) : ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5)

= ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5)

Add a constant to the solution

= ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5) + C

\frac{d}{d x} (arctan (e^{- x}))

e^{3 x}

n \to \infty lim (\frac{n ^{2}}{2 1 ^{n}})

5 t^{2} - 2 e^{t}

\frac{\partial}{\partial y} (4 x^{3} - 4 y)

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