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

The integral of ((t-1))/(sqrt(t^2-2t)) is sqrt(t^2-2t)+C

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

\int \frac{( t - 1 )}{t ^{2} - 2 t} d t

t^{2} - 2 t + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = t^{2} - 2 t

= \frac{1}{2} \cdot 2 (t^{2} - 2 t)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (t^{2} - 2 t)^{\frac{1}{2}} : t^{2} - 2 t

= t^{2} - 2 t

Add a constant to the solution

= t^{2} - 2 t + C

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