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

The integral of (5-4x)/(sqrt(x^2-3x+2)) is -4sqrt(x^2-3x+2)+ln(1/2 |-4x+4sqrt(x^2-3x+2)+6|)+C

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

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

- 4 x^{2} - 3 x + 2 + ln (\frac{1}{2} - 4 x + 4 x^{2} - 3 x + 2 + 6) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Complete the square

u^{2} + 2 u - 3 : (u + 1)^{2} - 4

= - \int \frac{u}{( u + 1 ) ^{2} - 4} d u

Apply u-substitution

= - \int \frac{v - 1}{v ^{2} - 4} d v

Expand

\frac{v - 1}{v ^{2} - 4} : \frac{v}{v ^{2} - 4} - \frac{1}{v ^{2} - 4}

= - \int \frac{v}{v ^{2} - 4} - \frac{1}{v ^{2} - 4} 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

= - (\int \frac{v}{v ^{2} - 4} d v - \int \frac{1}{v ^{2} - 4} d v)

\int \frac{v}{v ^{2} - 4} d v = v^{2} - 4

\int \frac{1}{v ^{2} - 4} d v = ln (\frac{1}{2} v^{2} - 4 + v)

= - (v^{2} - 4 - ln (\frac{1}{2} v^{2} - 4 + v))

Substitute back

= - ((5 - 4 x + 1)^{2} - 4 - ln (\frac{1}{2} (5 - 4 x + 1)^{2} - 4 + 5 - 4 x + 1))

Simplify

- ((5 - 4 x + 1)^{2} - 4 - ln (\frac{1}{2} (5 - 4 x + 1)^{2} - 4 + 5 - 4 x + 1)) : - 4 x^{2} - 3 x + 2 + ln (\frac{1}{2} - 4 x + 4 x^{2} - 3 x + 2 + 6)

= - 4 x^{2} - 3 x + 2 + ln (\frac{1}{2} - 4 x + 4 x^{2} - 3 x + 2 + 6)

Add a constant to the solution

= - 4 x^{2} - 3 x + 2 + ln (\frac{1}{2} - 4 x + 4 x^{2} - 3 x + 2 + 6) + C

What is the integral of (5-4x)/(sqrt(x^2-3x+2)) ?
The integral of (5-4x)/(sqrt(x^2-3x+2)) is -4sqrt(x^2-3x+2)+ln(1/2 |-4x+4sqrt(x^2-3x+2)+6|)+C