Answer to Question #101125 in Calculus for Jason

Question

Integrate with quotient development as needed. Tip: In a)
first, divide by, for example, a split angle.

a) ∫2x2-x+2/x-1dx

d) ∫x-1/x2-4dx

Answer 1

a) Let us rewrite the numerator as follows: $2x^2 - x + 2 = 2[ x^2 - \frac{x}{2} + 1] = 2[ (x-1)^2 + \frac{3x}{2}]$ - we have completed the square in the second term, to cancel the term $(x-1)^2$ with the denominator.

Hence, $\int \frac{2 x^2 - x + 2}{x-1} = 2 \int (x-1) dx + 3 \int \frac{x}{x-1} dx = 2 (\frac{x^2}{2} - x) + 3 \int \frac{(x-1) +1}{x-1} dx = x^2 - 2x + 3x + 3 \int \frac{1}{x-1} dx = x^2 +x + 3 \ln(x-1) + C$

b) Let us decompose the integrand into partial fractions: $\frac{x-1}{x^2-4} = \frac{x-1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2} = \frac{(A+B)x + (2A-2B)}{(x-2)(x+2)}$, from where comparing the numerator of right and left side, obtain linear system: $A + B = 1; 2A - 2 B = 1$, which has the solution $A = \frac{1}{4}, B = \frac{3}{4}$.

Hence, $\int \frac{x-1}{x^2-4} dx = \int \frac{1}{4(x-2)}dx + \int \frac{3}{4(x+2)}dx = \frac{\ln(x-2)}{4} + \frac{3 \ln(x+2)}{4} + C$, where we used table integral $\int \frac{1}{x}dx = \ln x + C$.