long division\nwhen integrating a fraction where the degree of the numerator ≥ the degree of the…

long division\nwhen integrating a fraction where the degree of the numerator ≥ the degree of the denominator, you w\nto use long division (or creative thinking) to “split the fraction.”\nexample: integrate $\\int\\frac{x^{2}-4x + 2}{x^{2}+2}dx$\nassigned problems:\n1. $\\int\\frac{x + 1}{x - 1}dx$ 2. $\\int\\frac{x^{3}}{x^{2}+1}dx$ 3. $\\int_{0}^{1}\\frac{x^{2}}{x + 1}dx$\n4. $\\int\\frac{5x^{2}}{x^{2}+1}dx$ 5. $\\int\\frac{2x^{2}-4}{x + 1}dx$ 6. $\\int_{2}^{3}\\frac{y + 1}{y - 1}dy$

long division\nwhen integrating a fraction where the degree of the numerator ≥ the degree of the denominator, you w\nto use long division (or creative thinking) to “split the fraction.”\nexample: integrate $\\int\\frac{x^{2}-4x + 2}{x^{2}+2}dx$\nassigned problems:\n1. $\\int\\frac{x + 1}{x - 1}dx$ 2. $\\int\\frac{x^{3}}{x^{2}+1}dx$ 3. $\\int_{0}^{1}\\frac{x^{2}}{x + 1}dx$\n4. $\\int\\frac{5x^{2}}{x^{2}+1}dx$ 5. $\\int\\frac{2x^{2}-4}{x + 1}dx$ 6. $\\int_{2}^{3}\\frac{y + 1}{y - 1}dy$

Answer

Explanation:

Step1: Use long - division for (\frac{x + 1}{x-1})

[ \begin{align*} \frac{x + 1}{x-1}&=\frac{(x - 1)+2}{x-1}\ &=1+\frac{2}{x - 1} \end{align*} ]

Step2: Integrate term - by - term

[ \begin{align*} \int\frac{x + 1}{x-1}dx&=\int\left(1+\frac{2}{x - 1}\right)dx\ &=\int 1dx+2\int\frac{1}{x - 1}dx \end{align*} ] Using the integral formulas (\int kdx=kx+C) ((k = 1) here) and (\int\frac{1}{u}du=\ln|u|+C) ((u=x - 1), (du=dx)) [ \begin{align*} \int 1dx+2\int\frac{1}{x - 1}dx&=x+2\ln|x - 1|+C \end{align*} ]

Answer:

(x + 2\ln|x - 1|+C)