Question

$\int \frac{1}{x \left(6(\log x)^2 + 7\log x + 2\right)} \, dx =$

• A. $\frac{1}{2} \frac{\log \left|2\log x + 1\right|}{3\log x + 2} + C$
• B. $\log \frac{\left|2\log x + 1\right|}{3\log x + 2} + C$
• C. $\log \frac{\left|3\log x + 2\right|}{2\log x + 1} + C$
• D. $\frac{1}{2} \frac{\log \left|3\log x + 2\right|}{2\log x + 1} + C$

Answer 1

The Correct Option is B

1. Understand the integral:

We need to evaluate \( \int \frac{1}{6(\log x)^2 + 7\log x + 2} \, dx \).

2. Use substitution:

Let \( u = \log x \), then \( du = \frac{1}{x} dx \) or \( x du = dx \).

The integral becomes:

\[ \int \frac{x}{6u^2 + 7u + 2} du \]

But since \( x = e^u \), we have:

\[ \int \frac{e^u}{6u^2 + 7u + 2} du \]

3. Factor the denominator:

\[ 6u^2 + 7u + 2 = (2u + 1)(3u + 2) \]

4. Apply partial fractions:

\[ \frac{1}{(2u + 1)(3u + 2)} = \frac{A}{2u + 1} + \frac{B}{3u + 2} \]

Solving for \( A \) and \( B \):

\[ 1 = A(3u + 2) + B(2u + 1) \]

Setting \( u = -\frac{1}{2} \): \( 1 = A\left(-\frac{3}{2} + 2\right) \implies A = 2 \)

Setting \( u = -\frac{2}{3} \): \( 1 = B\left(-\frac{4}{3} + 1\right) \implies B = -3 \)

Thus:

\[ \frac{1}{(2u + 1)(3u + 2)} = \frac{2}{2u + 1} - \frac{3}{3u + 2} \]

5. Integrate:

\[ \int \left( \frac{2}{2u + 1} - \frac{3}{3u + 2} \right) du = \log |2u + 1| - \log |3u + 2| + C = \log \left| \frac{2u + 1}{3u + 2} \right| + C \]

6. Substitute back \( u = \log x \):

\[ \log \left| \frac{2\log x + 1}{3\log x + 2} \right| + C \]

Correct Answer: (B) \( \log \left| \frac{2\log x + 1}{3\log x + 2} \right| + C \)

Answer 2

Substitute $u = \log x$, so $du = \frac{1}{x} dx$. The integral becomes: \[ I = \int \frac{1}{6u^2 + 7u + 2} \, du. \] Factorize $6u^2 + 7u + 2$: \[ 6u^2 + 7u + 2 = (3u + 2)(2u + 1). \] Use partial fractions: \[ \frac{1}{(3u + 2)(2u + 1)} = \frac{A}{3u + 2} + \frac{B}{2u + 1}. \] Solve for $A$ and $B$, and integrate: \[ I = \log \frac{\left|2\log x + 1\right|}{3\log x + 2} + C. \]