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Question
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
Sum
Solution
\[\int\frac{1}{x^2 + 4x - 5}dx\]
\[ = \int\frac{1}{x^2 + 4x + 4 - 4 - 5}dx\]
\[ = \int\frac{1}{x^2 + 4x + 4 - 3^2}dx\]
\[ = \int\frac{1}{\left( x + 2 \right)^2 - 3^2}dx\]
\[ = \frac{1}{2 \times 3} \text{ ln} \left| \frac{x + 2 - 3}{x + 2 + 3} \right| + C ................. \left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]
\[ = \frac{1}{6} \text{ ln } \left| \frac{x - 1}{x + 5} \right| + C\]
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