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

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

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

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

x - 2 x - 1 - ln x - 2 + x - 1 + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int tan^{2} (v) sec (v) d v

= 2 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 2 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= 2 \cdot \int - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int sec (v) d v + \int sec^{3} (v) d v)

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

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= 2 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

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

Simplify

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

= x - 2 x - 1 - ln x - 2 + x - 1

Add a constant to the solution

= x - 2 x - 1 - ln x - 2 + x - 1 + C

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