Question

If \(\int \frac{dx}{(x+1)(x-2)(x-3)}=\frac{1}{k}log_e\left \{ \frac{|x-3|^3|x+1|}{(x-2)^4}\right \}+c\), then the value of k is 

• A. 4
• B. 6
• C. 8
• D. 12

Answer 1

The Correct Option is D

Step 1: Partial Fraction Decomposition

We need to decompose the integrand into partial fractions:

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

Multiply both sides by \((x+1)(x-2)(x-3)\):

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

Step 2: Solve for A, B, and C

* For \(x = -1\):

\(1 = A(-1-2)(-1-3) = A(-3)(-4) = 12A \implies A = \frac{1}{12}\)

* For \(x = 2\):

\(1 = B(2+1)(2-3) = B(3)(-1) = -3B \implies B = -\frac{1}{3}\)

* For \(x = 3\):

\(1 = C(3+1)(3-2) = C(4)(1) = 4C \implies C = \frac{1}{4}\)

Step 3: Rewrite the Integral

Now we can rewrite the integral as:

\(I = \int \left( \frac{1/12}{x+1} - \frac{1/3}{x-2} + \frac{1/4}{x-3} \right) dx\)

\(I = \frac{1}{12} \int \frac{1}{x+1} dx - \frac{1}{3} \int \frac{1}{x-2} dx + \frac{1}{4} \int \frac{1}{x-3} dx\)

Step 4: Evaluate the Integrals

\(I = \frac{1}{12} \log |x+1| - \frac{1}{3} \log |x-2| + \frac{1}{4} \log |x-3| + C\)

Step 5: Combine Logarithms

\(I = \log \left| (x+1)^{1/12} \right| - \log \left| (x-2)^{1/3} \right| + \log \left| (x-3)^{1/4} \right| + C\)

\(I = \log \left| \frac{(x+1)^{1/12} (x-3)^{1/4}}{(x-2)^{1/3}} \right| + C\)

Now, let's rewrite to match the given form:

\(I = \frac{1}{k} \log_e \left| \frac{(x-3)^3 (x+1)}{(x-2)^4} \right| + C\)

\(I = \frac{1}{12}log|x+1| - \frac{1}{3}log|x-2|+\frac{1}{4}log|x-3| \)

\(I = \frac{1}{12} (log|x+1|-4log|x-2|+3log|x-3|) \) \(I = \frac{1}{12} (log|x+1|+log|x-3|^3-log|x-2|^4) \) \(I = \frac{1}{12} log \frac{|x+1||x-3|^3}{|x-2|^4} \)

By direct comparison \(k=12 \)

Conclusion:

The value of k is 12.