find a reduction formulae for I_n = ∫_(−1) ^0 x^n (1 + x)^2 dx


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Question Number 93299 by Rio Michael last updated on 12/May/20

$find a reduction formulae for I_{n} = \int_{- 1}^{0} x^{n} {(1 + x)}^{2} d x$
Commented by mathmax by abdo last updated on 12/May/20
$I_n =∫_(−1) ^0 x^n (1+x)^2 dx ⇒I_n =_(x=−t) ∫_0 ^1 (−1)^n t^n (1−t)^2 dt =(−1)^n ∫_0 ^1 t^n (t−1)^2 dt by parts we get ∫_0 ^1 t^n (t−1)^2 dt =[(1/(n+1))t^(n+1) (t−1)^2 ]_0 ^1 −(2/(n+1))∫_0 ^1 t^(n+1) (t−1)dt =−(2/(n+1)) { [(1/(n+2))t^(n+2) (t−1)]_0 ^1 −(1/(n+2))∫_0 ^1 t^(n+2) dt} =−(2/(n+1))(−(1/(n+2)))×(1/(n+3)) =(2/((n+1)(n+2)(n+3))) ⇒ I_n =((2(−1)^n )/((n+1)(n+2)(n+3)))$
$I_{n} = \int_{- 1}^{0} x^{n} {(1 + x)}^{2} d x \Rightarrow I_{n} =_{x = - t} \int_{0}^{1} {(- 1)}^{n} t^{n} {(1 - t)}^{2} d t$ $= {(- 1)}^{n} \int_{0}^{1} t^{n} {(t - 1)}^{2} d t b y p a r t s w e g e t$ $\int_{0}^{1} t^{n} {(t - 1)}^{2} d t = {[\frac{1}{n + 1} t^{n + 1} {(t - 1)}^{2}]}_{0}^{1} - \frac{2}{n + 1} \int_{0}^{1} t^{n + 1} (t - 1) d t$ $= - \frac{2}{n + 1} {{[\frac{1}{n + 2} t^{n + 2} (t - 1)]}_{0}^{1} - \frac{1}{n + 2} \int_{0}^{1} t^{n + 2} d t}$ $= - \frac{2}{n + 1} (- \frac{1}{n + 2}) \times \frac{1}{n + 3} = \frac{2}{(n + 1) (n + 2) (n + 3)} \Rightarrow$ $I_{n} = \frac{2 {(- 1)}^{n}}{(n + 1) (n + 2) (n + 3)}$
Commented by Rio Michael last updated on 12/May/20

$thank you sir$
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