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Question
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
Solution
Let \[I=\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
Put 2x + 1 = z2
\[\Rightarrow 2dx = 2zdz\]
\[ \Rightarrow dx = zdz\]
When
\[x \to 2, z \to \sqrt{5}\]
When
\[x \to 4, z \to 3\]
\[\therefore I = \int_\sqrt{5}^3 \frac{\left( \frac{z^2 - 1}{2} \right)^2 + \frac{z^2 - 1}{2}}{z} \times zdz\]
\[ \Rightarrow I = \int_\sqrt{5}^3 \frac{z^4 - 2 z^2 + 1 + 2 z^2 - 2}{4}dz\]
\[ \Rightarrow I = \frac{1}{4} \int_\sqrt{5}^3 \left( z^4 - 1 \right)dz\]
\[ \Rightarrow I = \left.\frac{1}{4} \times \left( \frac{z^5}{5} - z \right)\right|_\sqrt{5}^3\]
\[\Rightarrow I = \frac{1}{4}\left[ \left( \frac{243}{5} - 3 \right) - \left( \frac{25\sqrt{5}}{5} - \sqrt{5} \right) \right]\]
\[ \Rightarrow I = \frac{1}{4} \times \frac{228}{5} - \frac{1}{4} \times 4\sqrt{5}\]
\[ \Rightarrow I = \frac{57}{5} - \sqrt{5}\]
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