What is the integral of (3x-6)/(sqrt(x^2-4x+5)) ?

The integral of (3x-6)/(sqrt(x^2-4x+5)) is 3sqrt(x^2-4x+5)+C

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integral of (3x-6)/(sqrt(x^2-4x+5))

\int \frac{3 x - 6}{x ^{2} - 4 x + 5} d x

3 x^{2} - 4 x + 5 + C

Solution steps

\int \frac{3 x - 6}{x ^{2} - 4 x + 5} d x

Complete the square

x^{2} - 4 x + 5 : (x - 2)^{2} + 1

= \int \frac{3 x - 6}{( x - 2 ) ^{2} + 1} d x

Apply u-substitution

= \int \frac{3 u}{u ^{2} + 1} d u

Take the constant out:

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

= 3 \cdot \int \frac{u}{u ^{2} + 1} d u

Apply u-substitution

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Apply the Power Rule

= 3 \cdot \frac{1}{2} \cdot 2 v^{\frac{1}{2}}

Substitute back

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

Simplify

3 \cdot \frac{1}{2} \cdot 2 ((x - 2)^{2} + 1)^{\frac{1}{2}} : 3 x^{2} - 4 x + 5

= 3 x^{2} - 4 x + 5

Add a constant to the solution

= 3 x^{2} - 4 x + 5 + C

\frac{\partial}{\partial x} (x z x + y)

\int - 5 x^{4} e^{x^{5}} d x

d er i v a t i v e f (x) = \frac{x}{32} (x + 3 x)

\frac{\partial}{\partial y} (sin (x^{2} y))

\frac{d}{d x} ((x^{3} + 6 x)^{4})

What is the integral of (3x-6)/(sqrt(x^2-4x+5)) ?
The integral of (3x-6)/(sqrt(x^2-4x+5)) is 3sqrt(x^2-4x+5)+C