Question

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

Answer 1

We have,

\[I = \int\left( \frac{x^2 + 6x - 8}{x^3 - 4x} \right)dx\]

\[ = \int\frac{\left( x^2 + 6x - 8 \right)}{x \left( x^2 - 4 \right)}dx\]

\[ = \int\frac{\left( x^2 + 6x - 8 \right)}{x \left( x - 2 \right) \left( x + 2 \right)}dx\]

\[Let \frac{x^2 + 6x - 8}{x \left( x - 2 \right) \left( x + 2 \right)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}\]

\[ \Rightarrow \frac{x^2 + 6x - 8}{x \left( x - 2 \right) \left( x + 2 \right)} = \frac{A \left( x - 2 \right) \left( x + 2 \right) + B \left( x \right) \left( x + 2 \right) + C \left( x \right) \left( x - 2 \right)}{x \left( x - 2 \right) \left( x + 2 \right)}\]

\[ \Rightarrow x^2 + 6x - 8 = A \left( x^2 - 4 \right) + B \left( x^2 + 2x \right) + C \left( x^2 - 2x \right)\]

Putting `x - 2 = 0`

\[ \Rightarrow x = 2\]

\[4 + 6 \times 2 - 8 = A \times 0 + B \left( 4 + 4 \right)\]

\[ \Rightarrow 8 = B \times 8\]

\[ \Rightarrow B = 1\]

Putting `x = - 2`

\[4 - 12 - 8 = A \times 0 + B \times 0 + C \times 8\]
\[ \Rightarrow C = - 2\]

Putting `x = 0`

\[ - 8 = A \left( - 4 \right) + B \times 0 + C \times 0\]

\[ \Rightarrow A = 2\]

\[ \therefore I = \int\frac{2}{x} + \int\frac{dx}{x - 2} - 2\int\frac{dx}{x + 2}\]

\[ = 2 \log \left| x \right| + \log \left| x - 2 \right| - 2 \log \left| x + 2 \right| + C\]

\[ = \log x^2 + \log \left| x - 2 \right| - \log \left| x + 2 \right|^2 + C\]

\[ = \log \left| \frac{x^2 \left( x - 2 \right)}{\left( x + 2 \right)^2} \right| + C\]