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Question Number 216918 by mathocean1 last updated on 24/Feb/25

$$givearecurrencerelationfor{I}_n.$$$$I_n={\int }_0^1\frac{x^n}{x+3}dx,\mathrm{\forall }n\in \mathbb{N}.$$

Answered by Wuji last updated on 24/Feb/25

$${\mathrm{I}}_{\mathrm{n}}=\underset{0}{\stackrel{1}{\int }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{x}+3}\mathrm{dx},\mathrm{\forall }\mathrm{n}\in \mathbb{N}$$$$({\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{n}-1}\mathrm{x}={\mathrm{x}}^{\mathrm{n}-1}((\mathrm{x}+3)-3)$$$$\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{x}+3}=\frac{{\mathrm{x}}^{\mathrm{n}-1}((\mathrm{x}+3)-3)}{(\mathrm{x}+3)}={\mathrm{x}}^{\mathrm{n}-1}-3\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{x}+3}$$$${\mathrm{I}}_{\mathrm{n}}=\underset{0}{\stackrel{1}{\int }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{x}+3}\mathrm{dx}=\underset{0}{\stackrel{1}{\int }}({\mathrm{x}}^{\mathrm{n}-1}-3\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{x}+3})\mathrm{dx}$$$$\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^{\mathrm{n}-1}\mathrm{dx}=\frac{1}{\mathrm{n}},\underset{0}{\stackrel{1}{\int }}\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{x}-3}\mathrm{dx}={\mathrm{I}}_{\mathrm{n}-1}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{1}{\mathrm{n}}-{3\mathrm{I}}_{\mathrm{n}-1}$$

Commented by mathocean1 last updated on 24/Feb/25

$$Thankyousir...$$

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