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

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

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

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

2 ln x + x - 1 + C

Solution steps

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

Factor

\frac{1}{x ^{2} - x}

= \int \frac{1}{x ( x - 1 )} d x

Apply u-substitution

: \int \frac{1}{( u + 1 ) u} d u

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

(u + 1) u = (u + 1) u

= \int \frac{1}{u + 1 u} d u

Apply u-substitution

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

= \int \frac{2}{v ^{2} + 1} d v

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = ln v^{2} + 1 + v

= 2 ln v^{2} + 1 + v

Substitute back

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

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

= 2 ln x + x - 1

Add a constant to the solution

= 2 ln x + x - 1 + C

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cos (6 x), 0

y^{^{''}} + 43 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 0

n = 1 \sum \infty \frac{2}{n + 1}

\int_{0}^{π} x + sin (x) d x

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