Question

Integrate with respect to x: ∫ x^2/((x-2)(x^2+1)) dx

4/5 ln⁡(x-2)+2/5 tan^(-1) x+c

4/5 ln(x-2)+1/5 ln⁡(x^2+1)+2/5 tan^(-1) x+c

4/5 ln(x-2)+1/10 ln⁡(x^2+1)+2/5 tan^(-1) x+c

4/5 ln(x-3)+1/10 ln⁡(x^2+1)+2/5 tan^(-1) x+c

Answer 1

Integrate with respect to x: $\int x^2 / (x - 2)(x^2 + 1) \, dx$

$$\begin{array}{l} 4/5 \ln(\pi(x - 2) + 2/5 \tan^(-1) x + c \\ 4/5 \ln(x - 2) + 1/5 \ln(\pi(x^2 + 1) + 2/5 \tan^(-1) x + c \\ \end{array}$$

Solution

$$\begin{array}{l} I = \int \frac{x^2}{(x - 2)(x^2 + 1)} \, dx = \int \frac{x^2 + 1 - 1}{(x - 2)(x^2 + 1)} \, dx = \int \left(\frac{1}{(x - 2)} - \frac{1}{(x - 2)(x^2 + 1)}\right) dx \\ \frac{1}{(x - 2)(x^2 + 1)} = \frac{A}{(x - 2)} + \frac{Bx + C}{(x^2 + 1)} \\ \end{array}$$

Let's find A,B,C:

$$\begin{array}{l} x^2: A + B = 0 \rightarrow B = -A \\ x: C - 2B = 0 \rightarrow C = 2B = -2A \\ 1: A - 2C = 1 \rightarrow A - 2(-2A) = 1 \rightarrow A = \frac{1}{5}, B = -\frac{1}{5}, C = -\frac{2}{5}. \\ \end{array}$$

So

$$\begin{array}{l} \frac{1}{(x - 2)(x^2 + 1)} = \frac{1}{5} \left(\frac{1}{(x - 2)} - \frac{x + 2}{(x^2 + 1)}\right). \\ I = \int \left(\frac{4}{5} \frac{1}{(x - 2)} + \frac{1}{5} \frac{x + 2}{(x^2 + 1)}\right) dx = I_1 + I_2 + c \\ I_1 = \int \left(\frac{4}{5} \frac{1}{(x - 2)}\right) dx = \frac{4}{5} \int \left(\frac{1}{(x - 2)}\right) dx = \frac{4}{5} \int \frac{d(x - 2)}{(x - 2)} = \frac{4}{5} \ln(x - 2). \\ I_2 = \int \left(\frac{1}{5} \frac{x + 2}{(x^2 + 1)}\right) dx = \frac{1}{5} \int \frac{x + 2}{x^2 + 1} dx = \frac{1}{5} \int \left(\frac{x}{x^2 + 1} + \frac{2}{x^2 + 1}\right) dx = I_3 + I_4 \\ I_3 = \frac{1}{5} \int \left(\frac{x}{x^2 + 1}\right) dx = \frac{1}{5} \int \left(\frac{d\left(\frac{x^2}{2}\right)}{x^2 + 1}\right) = \frac{1}{10} \int \frac{d(x^2 + 1)}{x^2 + 1} = \frac{1}{10} \ln(x^2 + 1) \\ I_4 = \frac{1}{5} \int \left(\frac{2}{x^2 + 1}\right) dx = \frac{2}{5} \int \frac{dx}{x^2 + 1} = \frac{2}{5} \tan^{-1} x. \\ \end{array}$$

Finally we have

$$I = \frac{4}{5} \ln(x - 2) + \frac{1}{10} \ln(x^2 + 1) + \frac{2}{5} \tan^{-1} x + c.$$

Answer: $\frac{4}{5} \ln(x - 2) + \frac{1}{10} \ln(x^2 + 1) + \frac{2}{5} \tan^{-1} x + c.$