Evaluate: \(\rm \int \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx\)1. \(\rm log

1.	Evaluate: \(\rm \int \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx\)1. \(\rm log\left \| x+3 \right \|+C\)2. \(\rm 10log\left \| x+3 \right \|+C\)3. \(\rm 3log\left \| x+1 \right \|-12log\left \| x+2 \right \|+10log\left \| x+3 \right \|+C\)4. \(\rm 3log\left \| x+1 \right \|+12log\left \| x+2 \right \|+10log\left \| x+3 \right \|+C\)
Answer» Correct Answer - Option 3 : \(\rm 3log\left \| x+1 \right \|-12log\left \| x+2 \right \|+10log\left \| x+3 \right \|+C\) Concept: \(\rm \int\frac{1}{x+a}dx=log\|x+a\|+C\) \(\rm \int\frac{1}{(x+a)^n}dx=\frac{1}{(1-n)(x+a)^{n+1}}+C\) Calculation: To solve: \(\rm \int \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx\) ⇒\(\rm \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\) ⇒ x² - 3x + 2 = A(x + 2)(x + 3) + B(x + 1)(x + 3) +C(x + 1)(x + 2) ⇒ x2 - 3x + 2 = (A + B +C)x2 + (5A + 4B + 3C)x + (6A + 3B + 2C) By comparing coefficient of x2, x and constant terms on both sides, we get A + B +C = 1, 5A + 4B + 3C = -3 and 6A + 3B + 2C = 2 By solving these equation, we get A = 3, B = -12 and C = 10 ⇒\(\rm \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}=\frac{3}{x+1}-\frac{12}{x+2}+\frac{10}{x+3}\) ⇒\(\rm \int\frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx=\int\frac{3}{x+1}dx-\int\frac{12}{x+2}dx+\int\frac{10}{x+3}dx\) ⇒ \(\rm \int\frac{x^3-3x+2}{(x+1)(x+2)(x+3)}dx=3log\left \| x+1 \right \|-12log\left \| x+2 \right \|+10log\left \| x+3 \right \|+C\) Hence, option 3 is correct.

Evaluate: \(\rm \int \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx\)1. \(\rm log\left | x+3 \right |+C\)2. \(\rm 10log\left | x+3 \right |+C\)3. \(\rm 3log\left | x+1 \right |-12log\left | x+2 \right |+10log\left | x+3 \right |+C\)4. \(\rm 3log\left | x+1 \right |+12log\left | x+2 \right |+10log\left | x+3 \right |+C\)

Concept:

Calculation:

To solve: \(\rm \int \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx\)

⇒\(\rm \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\)

⇒ x² - 3x + 2 = A(x + 2)(x + 3) + B(x + 1)(x + 3) +C(x + 1)(x + 2)

⇒ x2 - 3x + 2 = (A + B +C)x2 + (5A + 4B + 3C)x + (6A + 3B + 2C)

By comparing coefficient of x2, x and constant terms on both sides, we get A + B +C = 1, 5A + 4B + 3C = -3 and 6A + 3B + 2C = 2

By solving these equation, we get A = 3, B = -12 and C = 10

⇒\(\rm \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}=\frac{3}{x+1}-\frac{12}{x+2}+\frac{10}{x+3}\)

⇒\(\rm \int\frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx=\int\frac{3}{x+1}dx-\int\frac{12}{x+2}dx+\int\frac{10}{x+3}dx\)

⇒ \(\rm \int\frac{x^3-3x+2}{(x+1)(x+2)(x+3)}dx=3log\left | x+1 \right |-12log\left | x+2 \right |+10log\left | x+3 \right |+C\)

Hence, option 3 is correct.