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

The integral of (x^4+3)/(x^2-4x+3) is (x^3)/3+2x^2+13x-2ln|x-1|+42ln|x-3|+C

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

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

\frac{x ^{3}}{3} + 2 x^{2} + 13 x - 2 ln ∣ x - 1 ∣ + 42 ln ∣ x - 3 ∣ + C

Solution steps

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

Long division

\frac{x ^{4} + 3}{x ^{2} - 4 x + 3} : x^{2} + 4 x + 13 + \frac{40 x - 36}{x ^{2} - 4 x + 3}

= \int x^{2} + 4 x + 13 + \frac{40 x - 36}{x ^{2} - 4 x + 3} 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 x^{2} d x + \int 4 x d x + \int 13 d x + \int \frac{40 x - 36}{x ^{2} - 4 x + 3} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 4 x d x = 2 x^{2}

\int 13 d x = 13 x

\int \frac{40 x - 36}{x ^{2} - 4 x + 3} d x = - 2 ln ∣ x - 1 ∣ + 42 ln ∣ x - 3 ∣

= \frac{x ^{3}}{3} + 2 x^{2} + 13 x - 2 ln ∣ x - 1 ∣ + 42 ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + 2 x^{2} + 13 x - 2 ln ∣ x - 1 ∣ + 42 ln ∣ x - 3 ∣ + C

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What is the integral of (x^4+3)/(x^2-4x+3) ?
The integral of (x^4+3)/(x^2-4x+3) is (x^3)/3+2x^2+13x-2ln|x-1|+42ln|x-3|+C