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प्रश्न
Evaluate:
उत्तर
\[\text{ Let I }= \int\left( \frac{x^2 + 4x}{x^3 + 6 x^2 + 5} \right) dx\]
\[\text{ Let x}^3 + 6 x^2 + 5 = t\]
\[ \Rightarrow \left( 3 x^2 + 12x \right) dx = dt\]
\[ \Rightarrow \left( x^2 + 4x \right) dx = \frac{dt}{3}\]
\[\text{ Putting x}^3 + 6 x^2 + 5 = t \text{ and }\left( x^2 + 4x \right) dx = \frac{dt}{3}\]
\[ \therefore I = \frac{1}{3}\int\frac{dt}{t}\]
\[ = \frac{1}{3} \text{ ln } \left| t \right| + C\]
\[ = \frac{1}{3}\text{ ln }\left| x^3 + 6 x^2 + 5 \right| + C\]
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