Question

If 

\[\int\frac{3x+1}{(x-1)(x-2)(x-3)}dx\]

 

\[= A \;log |x - 1| + B \;log |x - 2| + C\; log |x - 3| + C\]

, then the values of A, B and C are respectively.

• A. 5, -7, -5
• B. 2, -7, -5
• C. 5, -7, 5
• D. 2, -7, 5

Answer 1

The Correct Option is D

We are given \(\int \frac{3x+1}{(x-1)(x-2)(x-3)} dx = A \log |x-1| + B \log |x-2| + C \log |x-3| + C\), and we need to find the values of A, B, and C.

We can use partial fraction decomposition to express the integrand as:

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

Multiplying both sides by \((x-1)(x-2)(x-3)\), we get:

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

To find A, set x = 1:

\(3(1)+1 = A(1-2)(1-3)\)

\(4 = A(-1)(-2)\)

\(4 = 2A\)

\(A = 2\)

To find B, set x = 2:

\(3(2)+1 = B(2-1)(2-3)\)

\(7 = B(1)(-1)\)

\(B = -7\)

To find C, set x = 3:

\(3(3)+1 = C(3-1)(3-2)\)

\(10 = C(2)(1)\)

\(C = 5\)

Thus, A = 2, B = -7, and C = 5.

Therefore, the correct option is (D) 2, -7, 5.

Answer 2

We have:

$$ \int \frac{3x+1}{(x-1)(x-2)(x-3)} dx = A \log|x-1| + B \log|x-2| + C \log|x-3| + C. $$

We can decompose the fraction into partial fractions:

$$ \frac{3x+1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}. $$

Multiply through by the denominator $ (x-1)(x-2)(x-3) $ to get:

$$ 3x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2). $$

To solve for $ A $, $ B $, and $ C $, substitute specific values of $ x $:

If $ x = 1 $:

$$ 3(1)+1 = A(1-2)(1-3) \implies 4 = A(-1)(-2) \implies 4 = 2A \implies A = 2. $$

If $ x = 2 $:

$$ 3(2)+1 = B(2-1)(2-3) \implies 7 = B(1)(-1) \implies B = -7. $$

If $ x = 3 $:

$$ 3(3)+1 = C(3-1)(3-2) \implies 10 = C(2)(1) \implies C = 5. $$

Thus, $ A = 2 $, $ B = -7 $, and $ C = 5 $.