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प्रश्न
\[\int\frac{1}{x^2 - 10x + 34} dx\]
बेरीज
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उत्तर
\[\int\frac{dx}{x^2 - 10x + 34}\]
\[ = \int\frac{dx}{x^2 - 10x + 25 - 25 + 34}\]
\[ = \int\frac{dx}{\left( x - 5 \right)^2 + 9}\]
\[ = \int\frac{dx}{\left( x - 5 \right)^2 + 3^2}\]
\[\text{ let x } - 5 = t\]
\[ \Rightarrow dx = dt\]
\[Now, \int\frac{dx}{\left( x - 5 \right)^2 + 3^2}\]
\[ = \int\frac{dt}{t^2 + 3^2}\]
\[ = \frac{1}{3} \tan^{- 1} \left( \frac{t}{3} \right) + C\]
\[ = \frac{1}{3} \tan^{- 1} \left( \frac{x - 5}{3} \right) + C\]
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