Question

\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]

Answer 1

We have,

\[I = \int\frac{dx}{x \left\{ 6 \left( \log x \right)^2 + 7 \log x + 2 \right\}}\]

Putting `log x = t`

\[ \Rightarrow \frac{1}{x} dx = dt\]

\[ \therefore I = \int\frac{dt}{6 t^2 + 7t + 2}\]

\[ = \int\frac{dt}{\left( 3t + 2 \right) \left( 2t + 1 \right)}\]

\[\text{Let }\frac{1}{\left( 3t + 2 \right) \left( 2t + 1 \right)} = \frac{A}{3t + 2} + \frac{B}{2t + 1}\]

\[ \Rightarrow \frac{1}{\left( 3t + 2 \right) \left( 2t + 1 \right)} = \frac{A \left( 2t + 1 \right) + B \left( 3t + 2 \right)}{\left( 3t + 2 \right) \left( 2t + 1 \right)}\]

\[ \Rightarrow 1 = A \left( 2t + 1 \right) + B \left( 3t + 2 \right)\]

Putting `2t + 1 = 0`

\[ \Rightarrow t = - \frac{1}{2}\]

\[1 = 0 + B \left( 3 \times - \frac{1}{2} + 2 \right)\]

\[ \Rightarrow 1 = B \left( \frac{1}{2} \right)\]

\[ \Rightarrow B = 2\]

Putting `3t + 2 = 0`

\[ \Rightarrow t = - \frac{2}{3}\]

\[1 = A \left( 2 \times - \frac{2}{3} + 1 \right) + 0\]

\[ \Rightarrow 1 = A \left( - \frac{4}{3} + 1 \right)\]

\[ \Rightarrow 1 = A \left( - \frac{1}{3} \right)\]

\[ \Rightarrow A = - 3\]

\[ \therefore I = \int\left( - \frac{3}{3t + 2} + \frac{2}{2t + 1} \right)dt\]

\[ = - 3 \frac{\log \left| 3t + 2 \right|}{3} + 2 \frac{\log \left| 2t + 1 \right|}{2} + C\]

\[ = - \log \left| 3t + 2 \right| + \log \left| 2t + 1 \right| + C\]

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

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