What is the integral of (x-3)/(sqrt(x^2-6x-1)) ?

The integral of (x-3)/(sqrt(x^2-6x-1)) is sqrt(x^2-6x-1)+C

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integral of (x-3)/(sqrt(x^2-6x-1))

\int \frac{x - 3}{x ^{2} - 6 x - 1} d x

x^{2} - 6 x - 1 + C

Solution steps

\int \frac{x - 3}{x ^{2} - 6 x - 1} d x

Apply u-substitution

= \int \frac{u}{u ^{2} - 10} d u

Apply u-substitution

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

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}{v} d v

Apply the Power Rule

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

Substitute back

= \frac{1}{2} \cdot 2 ((x - 3)^{2} - 10)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 ((x - 3)^{2} - 10)^{\frac{1}{2}} : x^{2} - 6 x - 1

= x^{2} - 6 x - 1

Add a constant to the solution

= x^{2} - 6 x - 1 + C

What is the integral of (x-3)/(sqrt(x^2-6x-1)) ?
The integral of (x-3)/(sqrt(x^2-6x-1)) is sqrt(x^2-6x-1)+C