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

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

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

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

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

Solution steps

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

\frac{x ^{2}}{x ^{2} - 4 x + 3} = \frac{4 x - 3}{x ^{2} - 4 x + 3} + 1

= \int \frac{4 x - 3}{x ^{2} - 4 x + 3} + 1 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{4 x - 3}{x ^{2} - 4 x + 3} d x + \int 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

x^{\frac{3}{2}}, 4

\frac{\partial}{\partial x} (- 210 (5 x - y)^{5})

\frac{d}{d x} (1 + c 1 - x^{2})

y = \frac{6 x}{x ^{2} + 1}

\int_{- 2}^{2} 1 d x

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