What is the integral from-1 to 0 of 1/(sqrt(x^2-2x)) ?

The integral from-1 to 0 of 1/(sqrt(x^2-2x)) is -ln(2+sqrt(3))

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integral from-1 to 0 of 1/(sqrt(x^2-2x))

\int_{- 1}^{0} \frac{1}{x ^{2} - 2 x} d x

- ln (2 + 3)

Decimal

- 1.31695 \dots

Solution steps

\int_{- 1}^{0} \frac{1}{x ^{2} - 2 x} d x

Complete the square

x^{2} - 2 x : (x - 1)^{2} - 1

= \int_{- 1}^{0} \frac{1}{( x - 1 ) ^{2} - 1} d x

Apply u-substitution

= \int_{- 2}^{- 1} \frac{1}{u ^{2} - 1} d u

Apply Trigonometric Substitution

= \int_{\frac{2 π}{3}}^{π} sec (v) d v

Use the common integral:

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

= [ln ∣ tan (v) + sec (v) ∣]_{\frac{2 π}{3}}^{π}

Compute the boundaries:

- ln (2 + 3)

= - ln (2 + 3)

What is the integral from-1 to 0 of 1/(sqrt(x^2-2x)) ?
The integral from-1 to 0 of 1/(sqrt(x^2-2x)) is -ln(2+sqrt(3))