What is the integral of (x^3-6x^2+11x-6)/(4x^3-28x^2+56x-32) ?

The integral of (x^3-6x^2+11x-6)/(4x^3-28x^2+56x-32) is 1/4 x+1/4 ln|x-4|+C

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integral of (x^3-6x^2+11x-6)/(4x^3-28x^2+56x-32)

\int \frac{x ^{3} - 6 x ^{2} + 11 x - 6}{4 x ^{3} - 28 x ^{2} + 56 x - 32} d x

\frac{1}{4} x + \frac{1}{4} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x ^{3} - 6 x ^{2} + 11 x - 6}{4 x ^{3} - 28 x ^{2} + 56 x - 32} d x

Long division

\frac{x ^{3} - 6 x ^{2} + 11 x - 6}{4 x ^{3} - 28 x ^{2} + 56 x - 32} : \frac{1}{4} + \frac{1}{4 ( x - 4 )}

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

\int \frac{1}{4} d x = \frac{1}{4} x

\int \frac{1}{4 ( x - 4 )} d x = \frac{1}{4} ln ∣ x - 4 ∣

= \frac{1}{4} x + \frac{1}{4} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{1}{4} x + \frac{1}{4} ln ∣ x - 4 ∣ + C

\int_{- \infty}^{0} e^{5 x} d x

\frac{\partial}{\partial x} (5 x ln (x^{2} + y))

x \to 0 lim (x^{2} - 6)

\int 2 sin (ln (x)) d x

\frac{f ( x )}{g ( x ) + 1}

What is the integral of (x^3-6x^2+11x-6)/(4x^3-28x^2+56x-32) ?
The integral of (x^3-6x^2+11x-6)/(4x^3-28x^2+56x-32) is 1/4 x+1/4 ln|x-4|+C