give-a-recurrence-relation-for-I-n-I-n-0-1-x-n-x-3-dx-n-N-

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

g i v e a r e c u r r e n c e r e l a t i o n f o r I_{n} .

I_{n} = \int_{0}^{1} \frac{x^{n}}{x + 3} d x, \forall n \in N .

Answered by Wuji last updated on 24/Feb/25

I_{n} = \underset{0}{\int^{1}} \frac{x^{n}}{x + 3} dx, \forall n \in N

(x^{n} = x^{n - 1} x = x^{n - 1} ((x + 3) - 3)

\frac{x^{n}}{x + 3} = \frac{x^{n - 1} ((x + 3) - 3)}{(x + 3)} = x^{n - 1} - 3 \frac{x^{n - 1}}{x + 3}

I_{n} = \underset{0}{\int^{1}} \frac{x^{n}}{x + 3} dx = \underset{0}{\int^{1}} (x^{n - 1} - 3 \frac{x^{n - 1}}{x + 3}) dx

\underset{0}{\int^{1}} x^{n - 1} dx = \frac{1}{n}, \underset{0}{\int^{1}} \frac{x^{n - 1}}{x - 3} dx = I_{n - 1}

I_{n} = \frac{1}{n} - {3 I}_{n - 1}

Commented by mathocean1 last updated on 24/Feb/25

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