Question

Calculate each of the integrals in Exercises 17–46. For some integrals you may need to use polynomial long division, partial fractions, factoring or expanding, or the method of completing the square.

$\int \frac{2x^4-8x^3+9x-4}{x^2-4x+1}dx$

Short answer

The value is$\frac{2x^3}{3}-2x+\frac{1}{2}\ln |x^2-4x+1|+C$

Step-by-step solution

Step 1. Given Information:

The given integral is $\int \frac{2x^4-8x^3+9x-4}{x^2-4x+1}dx$

We want to calculate each of the integrals.

Step 2. Calculation:

Using long division we get

$$\begin{gathered} \frac{2x^4-8x^3+9x-4}{x^2-4x+1}=2x^2-2+\frac{x-2}{x^2-4x+1} \\ Now \\ \int \frac{2x^4-8x^3+9x-4}{x^2-4x+1}dx=\int \left(2x^2-2+\frac{x-2}{x^2-4x+1}\right)dx \\ =\frac{2x^3}{3}-2x+\frac{1}{2}\ln |x^2-4x+1|+C \end{gathered}$$