Question

$\int\frac{y^{2}-3y+2}{y^{2}+y}dy$ is equal to

• A. $y+2\log|y|-4\log|1+y|+C$
• B. $y+2\log|y|-6\log|1+y|+C$
• C. $y+3\log|y|-6\log|1+y|+C$
• D. $y+2\log|y|+6\log|1+y|+C$
• E. $y+7\log|y|-6\log|1+y|+C$

Hint:

Math Tip: Whenever the degree of the numerator is $\ge$ the degree of the denominator in a rational algebraic integral, you must always divide the polynomials first before applying partial fractions.

Answer 1

The Correct Option is B

Concept:
Calculus - Integration using Partial Fractions.
Step 1: Perform polynomial division.
The degree of the numerator equals the degree of the denominator. Divide the terms: $$ \frac{y^2-3y+2}{y^2+y} = \frac{(y^2+y) - 4y + 2}{y^2+y} = 1 + \frac{-4y+2}{y(y+1)} $$
Step 2: Set up partial fractions.
Decompose the rational part into partial fractions: $$ \frac{-4y+2}{y(y+1)} = \frac{A}{y} + \frac{B}{y+1} $$ Multiply by $y(y+1)$: $$ -4y + 2 = A(y+1) + By $$
Step 3: Solve for constants A and B.

• Let $y = 0$: $2 = A(1) \implies A = 2$
• Let $y = -1$: $-4(-1) + 2 = B(-1) \implies 6 = -B \implies B = -6$

Substitute $A$ and $B$ back into the integral expression: $$ \int \left( 1 + \frac{2}{y} - \frac{6}{y+1} \right) dy $$
Step 4: Integrate each term.
Integrate using the standard rule $\int \frac{1}{x} dx = \log|x|$: $$ \int 1 \,dy + \int \frac{2}{y} \,dy - \int \frac{6}{y+1} \,dy $$ $$ = y + 2\log|y| - 6\log|y+1| + C $$