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

The integral of (5x^2+2)/(x^3-4x^2+5x) is 23/10 ln|x^2-4x+5|+2/5 ln|x|+54/5 arctan(x-2)+C

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

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

\frac{23}{10} ln x^{2} - 4 x + 5 + \frac{2}{5} ln ∣ x ∣ + \frac{54}{5} arctan (x - 2) + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x ^{3} - 4 x ^{2} + 5 x} d x = 2 (\frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} - 4 x + 5 + 4 arctan (x - 2)))

= 5 (\frac{1}{2} ln x^{2} - 4 x + 5 + 2 arctan (x - 2)) + 2 (\frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} - 4 x + 5 + 4 arctan (x - 2)))

Simplify

5 (\frac{1}{2} ln x^{2} - 4 x + 5 + 2 arctan (x - 2)) + 2 (\frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} - 4 x + 5 + 4 arctan (x - 2))) : \frac{23}{10} ln x^{2} - 4 x + 5 + \frac{2}{5} ln ∣ x ∣ + \frac{54}{5} arctan (x - 2)

= \frac{23}{10} ln x^{2} - 4 x + 5 + \frac{2}{5} ln ∣ x ∣ + \frac{54}{5} arctan (x - 2)

Add a constant to the solution

= \frac{23}{10} ln x^{2} - 4 x + 5 + \frac{2}{5} ln ∣ x ∣ + \frac{54}{5} arctan (x - 2) + C

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What is the integral of (5x^2+2)/(x^3-4x^2+5x) ?
The integral of (5x^2+2)/(x^3-4x^2+5x) is 23/10 ln|x^2-4x+5|+2/5 ln|x|+54/5 arctan(x-2)+C