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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \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

x \to + 0 - lim (\frac{1}{x} \cdot + e^{\frac{1}{x}})

y = 6 x^{2} e^{x}

\int x^{2} (2 x + \frac{2}{3}) d x

y = (x^{3} - 9 x)^{12}, (- 3, 0)

\int (4 sin (x) + 2 cos (x)) d x

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