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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{5 ( x - 1 )} d x = \frac{1}{5} ln ∣ x - 1 ∣

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

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

Add a constant to the solution

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

\int x arctan (3 x) d x

x \to 0 - lim (\frac{∣ x - 7 ∣}{x - 7})

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 8

8 3 x^{2} + 9

\frac{\partial}{\partial x} (42 e^{7 x} y)

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