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Question
\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]
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Solution
\[Let\ I = \int_0^1 \frac{2x + 3}{5 x^2 + 1}\ d x . Then, \]
\[I = \int_0^1 \frac{2x}{5 x^2 + 1} d x + \int_0^1 \frac{3}{5 x^2 + 1} d x\]
\[ \Rightarrow I = \frac{1}{5} \int_0^1 \frac{10x}{5 x^2 + 1} d x + 3 \int_0^1 \frac{1}{\left( \sqrt{5}x \right)^2 + 1^2} d x\]
\[ \Rightarrow I = \frac{1}{5} \left[ \log \left( 5 x^2 + 1 \right) \right]_0^1 + \frac{3}{\sqrt{5}} \left[ \tan^{- 1} \left( \sqrt{5}x \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{- 1} \sqrt{5}\]
\[I = \int_0^1 \frac{2x}{5 x^2 + 1} d x + \int_0^1 \frac{3}{5 x^2 + 1} d x\]
\[ \Rightarrow I = \frac{1}{5} \int_0^1 \frac{10x}{5 x^2 + 1} d x + 3 \int_0^1 \frac{1}{\left( \sqrt{5}x \right)^2 + 1^2} d x\]
\[ \Rightarrow I = \frac{1}{5} \left[ \log \left( 5 x^2 + 1 \right) \right]_0^1 + \frac{3}{\sqrt{5}} \left[ \tan^{- 1} \left( \sqrt{5}x \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{- 1} \sqrt{5}\]
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