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Reduction Definition Math

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$\begin{aligned} I_n &= \int_{-1}^1 \frac{ x^n }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \int_{-1}^1 x^n \cdot \frac{ 1 }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \frac 12 \int_{-1}^1 x^{n-1} \cdot \frac{ (2x+3-3) }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \frac 12 \int_{-1}^1 x^{n-1} \sqrt{2x+3}\, \mathrm dx - \frac 32 \int_{-1}^1 x^{n-1} \cdot \frac 1{\sqrt{2x+3}}\, \mathrm dx \\ &= \frac 12 \left( \left[\underbrace{ \phantom- \frac{x^n}{n} \phantom- }_{\mathrm I}\cdot \underbrace{ \phantom- \sqrt{2x+3} \phantom- }_{\mathrm S} \right]_{-1}^1 - \int_{-1}^1 \underbrace{ \phantom- \frac{x^n}{n} \phantom- }_{\mathrm I}\cdot \underbrace{ \phantom- \frac{1}{2\sqrt{2x+3}} \cdot 2 \, \mathrm dx\phantom- }_{\mathrm D}\right) - \frac 32 \cdot I_{n-1} \\ &= \frac 12 \left( \frac 1n \cdot (\sqrt 5 - (-1)^n) - \frac 1n \cdot I_n \right) - \frac 32 \cdot I_{n-1} \\ &=\frac 1{2n} \cdot (\sqrt 5 - (-1)^n) - \frac 1{2n} \cdot I_n - \frac 32 \cdot I_{n-1}. \end{aligned}$