What is the integral of (3x^2-10)/(x^2-4x+4) ?

The integral of (3x^2-10)/(x^2-4x+4) is -(3x^2-10)/(x-2)+6(x-2+2ln|x-2|)+C

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integral of (3x^2-10)/(x^2-4x+4)

\int \frac{3 x ^{2} - 10}{x ^{2} - 4 x + 4} d x

- \frac{3 x ^{2} - 10}{x - 2} + 6 (x - 2 + 2 ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{3 x ^{2} - 10}{x ^{2} - 4 x + 4} d x

Apply Integration By Parts

: - \frac{3 x ^{2} - 10}{x - 2} - \int - \frac{6 x}{x - 2} d x

= - \frac{3 x ^{2} - 10}{x - 2} - \int - \frac{6 x}{x - 2} d x

\int - \frac{6 x}{x - 2} d x = - 6 (x - 2 + 2 ln ∣ x - 2 ∣)

= - \frac{3 x ^{2} - 10}{x - 2} - (- 6 (x - 2 + 2 ln ∣ x - 2 ∣))

Simplify

= - \frac{3 x ^{2} - 10}{x - 2} + 6 (x - 2 + 2 ln ∣ x - 2 ∣)

Add a constant to the solution

= - \frac{3 x ^{2} - 10}{x - 2} + 6 (x - 2 + 2 ln ∣ x - 2 ∣) + C

y^{^{''}} - 6 y^{^{'}} + 13 y = 0

\frac{d y}{d x} = x y^{3} (1 + x^{2})^{- \frac{1}{2}}, y (0) = 1

\frac{\partial}{\partial x} (- x y (- 15 - x - y))

\frac{\partial}{\partial x} (3 cos (3 x + 4 y))

\frac{d y}{d t} = \frac{t}{y - 2}, y (- 1) = 0

What is the integral of (3x^2-10)/(x^2-4x+4) ?
The integral of (3x^2-10)/(x^2-4x+4) is -(3x^2-10)/(x-2)+6(x-2+2ln|x-2|)+C