Question

hello Lovely ppl!!! Plz help me to solve this Question.....



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Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\int \frac{1}{ {4x}^{2} + 4x + 5 }dx \\ = \int \frac{1}{ {(2x + 1)}^{2} + {2}^{2} } dx \\ = > {(2x + 1)} = u \\ = > 2dx = du \\ = > dx = \frac{du}{2} \\ = > \int \frac{ \frac{1}{2} }{ {u}^{2} + 4 } du \\ = > \frac{1}{2} \ {tan}^{ - 1} ( \frac{u}{2} ) \\ = > \frac{1}{2} \ {tan}^{ - 1} ( \frac{2x + 1}{4} )$