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

The integral of (2x^3+3x^2-4)/(x^2-4x+3) is x^2+11x-13/2 ln|x-1|+65/2 ln|x-3|+6ln|x^2-4x+3|+C

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

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

x^{2} + 11 x - \frac{13}{2} ln ∣ x - 1 ∣ + \frac{65}{2} ln ∣ x - 3 ∣ + 6 ln x^{2} - 4 x + 3 + C

Solution steps

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

Expand

\frac{2 x ^{3} + 3 x ^{2} - 4}{x ^{2} - 4 x + 3} : \frac{2 x ^{3}}{x ^{2} - 4 x + 3} + \frac{3 x ^{2}}{x ^{2} - 4 x + 3} - \frac{4}{x ^{2} - 4 x + 3}

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

\int \frac{2 x ^{3}}{x ^{2} - 4 x + 3} d x = x^{2} + 8 x - ln ∣ x - 1 ∣ + 27 ln ∣ x - 3 ∣

\int \frac{3 x ^{2}}{x ^{2} - 4 x + 3} d x = 3 (2 ln x^{2} - 4 x + 3 - 5 (\frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x - 3 ∣) + x)

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

= x^{2} + 8 x - ln ∣ x - 1 ∣ + 27 ln ∣ x - 3 ∣ + 3 (2 ln x^{2} - 4 x + 3 - 5 (\frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x - 3 ∣) + x) - 4 (- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣)

Simplify

x^{2} + 8 x - ln ∣ x - 1 ∣ + 27 ln ∣ x - 3 ∣ + 3 (2 ln x^{2} - 4 x + 3 - 5 (\frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x - 3 ∣) + x) - 4 (- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣) : x^{2} + 11 x - \frac{13}{2} ln ∣ x - 1 ∣ + \frac{65}{2} ln ∣ x - 3 ∣ + 6 ln x^{2} - 4 x + 3

= x^{2} + 11 x - \frac{13}{2} ln ∣ x - 1 ∣ + \frac{65}{2} ln ∣ x - 3 ∣ + 6 ln x^{2} - 4 x + 3

Add a constant to the solution

= x^{2} + 11 x - \frac{13}{2} ln ∣ x - 1 ∣ + \frac{65}{2} ln ∣ x - 3 ∣ + 6 ln x^{2} - 4 x + 3 + C

\frac{d}{d z} (sin^{2} (z))

\frac{2}{s ^{2} + s - 2}

\frac{( s + 1 )}{s ^{2} + 2 s + 5}

\int \frac{6 t - 15}{3 t ^{2} - 2} d t

\int \frac{60 ln ( 3 ^{3^{x + 1}} )}{3 ^{x} - 1} d x

What is the integral of (2x^3+3x^2-4)/(x^2-4x+3) ?
The integral of (2x^3+3x^2-4)/(x^2-4x+3) is x^2+11x-13/2 ln|x-1|+65/2 ln|x-3|+6ln|x^2-4x+3|+C