∫(1+4x+x^2 )^m dx


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Question Number 63844 by mmkkmm000m last updated on 10/Jul/19

$\int {(1 + 4 x + x^{2})}^{m} d x$
Commented by mathmax by abdo last updated on 10/Jul/19
$let A_m =∫ (x^2 +4x+1)^m dx ⇒ A_m =∫ (x^2 +4x +4−3)^m dx =∫{(x+2)^2 −3}^m dx =∫ Σ_(k=0) ^m C_k ^m (x+2)^(2k) (−3)^(m−k) dx =Σ_(k=0) ^m (−3)^(m−k) C_m ^k ∫ (x+2)^(2k) dx =Σ_(k=0) ^m (−3)^(m−k) C_m ^k (1/(2k+1))(x+2)^(2k+1) +C =Σ_(k=0) ^m (−3)^(m−k) (C_m ^k /(2k+1))(x+2)^(2k+1) +C .$
$l e t A_{m} = \int {(x^{2} + 4 x + 1)}^{m} d x \Rightarrow$ $A_{m} = \int {(x^{2} + 4 x + 4 - 3)}^{m} d x = \int {{(x + 2)}^{2} - 3}^{m} d x$ $= \int \sum_{k = 0}^{m} C_{k}^{m} {(x + 2)}^{2 k} {(- 3)}^{m - k} d x$ $= \sum_{k = 0}^{m} {(- 3)}^{m - k} C_{m}^{k} \int {(x + 2)}^{2 k} d x$ $= \sum_{k = 0}^{m} {(- 3)}^{m - k} C_{m}^{k} \frac{1}{2 k + 1} {(x + 2)}^{2 k + 1} + C$ $= \sum_{k = 0}^{m} {(- 3)}^{m - k} \frac{C_{m}^{k}}{2 k + 1} {(x + 2)}^{2 k + 1} + C .$
Commented by mathmax by abdo last updated on 10/Jul/19

$a t l i n e 3 c h a n g e C_{k}^{m} b y C_{m}^{k}$
Commented by mathmax by abdo last updated on 10/Jul/19

$i h a v e t a k e n m f r o m N .$
	Answered by MJS last updated on 10/Jul/19

	${(a + b + c)}^{n} with n \in N$ $we need the t r i n o m i a l c o e f f i c i e n t s$ ${(a + b + c)}^{n} = \sum_{i, j, k} (\begin{matrix} n \\ i, j, k \end{matrix}) a^{i} b^{j} c^{k} with i + j + k = n$ $and (\begin{matrix} n \\ i, j, k \end{matrix}) = \frac{n!}{i! j! k!}$ $for m, i, j, k \in N \land i + j + k = m we get$ $\int {(1 + 4 x + x^{2})}^{m} d x =$ $= \int \sum_{i, j, k} (\begin{matrix} m \\ i, j, k \end{matrix}) 1^{i} {(4 x)}^{j} {(x^{2})}^{k} d x =$ $= \sum_{i, j, k} (\frac{m!}{i! j! k!} \int 4^{j} x^{j + 2 k} d x) =$ $= \sum_{i, j, k} (\frac{4^{j} m!}{(j + 2 k + 1) i! j! k!} x^{j + 2 k + 1}) + C$ $for m \in Z^{-} and m \in Q I^{'} m still trying . . .$
	Answered by Hope last updated on 10/Jul/19
	$∫{(x+2)^2 −3}^m dx y=(x+2) 3=a^2 I_m =∫(y^2 −a^2 )^m dy I_m =(y^2 −a^2 )^m y−m∫(y^2 −a^2 )^(m−1) ×2y×ydy =(y^2 −a^2 )^m ×y−2m∫(y^2 −a^2 )^(m−1) (y^2 −a^2 +a^2 )dy =(y^2 −a^2 )^m ×y−2mI_m −2ma^2 I_(m−1) I_m (1+2m)=y(y^2 −a^2 )^m −2ma^2 I_(m−1) I_m =((y(y^2 −a^2 )^m )/(1+2m))−((2ma^2 )/(1+2m))I_(m−1) ∫(1+4x+x^2 )^m dx =(((x+2)(1+4x+x^2 )^m )/(1+2m))−((2m×3)/(1+2m))∫(1+4x+x^2 )^(m−1) dx pls check...Tanmay$
	$\int {{(x + 2)}^{2} - 3}^{m} d x$ $y = (x + 2) 3 = a^{2}$ $I_{m} = \int {(y^{2} - a^{2})}^{m} d y$ $I_{m} = {(y^{2} - a^{2})}^{m} y - m \int {(y^{2} - a^{2})}^{m - 1} \times 2 y \times y d y$ $= {(y^{2} - a^{2})}^{m} \times y - 2 m \int {(y^{2} - a^{2})}^{m - 1} (y^{2} - a^{2} + a^{2}) d y$ $= {(y^{2} - a^{2})}^{m} \times y - 2 {m I}_{m} - 2 {m a}^{2} I_{m - 1}$ $I_{m} (1 + 2 m) = y {(y^{2} - a^{2})}^{m} - 2 {m a}^{2} I_{m - 1}$ $I_{m} = \frac{y {(y^{2} - a^{2})}^{m}}{1 + 2 m} - \frac{2 {m a}^{2}}{1 + 2 m} I_{m - 1}$ $\int {(1 + 4 x + x^{2})}^{m} d x$ $= \frac{(x + 2) {(1 + 4 x + x^{2})}^{m}}{1 + 2 m} - \frac{2 m \times 3}{1 + 2 m} \int {(1 + 4 x + x^{2})}^{m - 1} d x$ $p l s c h e c k . . . T a n m a y$
	Answered by MJS last updated on 10/Jul/19

	$\int {(1 + 4 x + x^{2})}^{m} d x with m \in Z^{-}$ $\int \frac{d x}{{(1 + 4 x + x^{2})}^{n}} with n \in N$ $trying {Ostrogradski}^{'} s Method$ $\int \frac{P (x)}{Q (x)} d x = \frac{P_{1} (x)}{Q_{1} (x)} + \int \frac{P_{2} (x)}{Q_{2} (x)} d x$ $Q_{1} (x) = \gcd (Q (x), Q^{'} (x))$ $Q_{2} (x) = \frac{Q (x)}{Q_{1} (x)}$ $P_{1} (x), P_{2} (x) can be found by comparing the$ $constant factors of$ $\frac{P (x)}{Q (x)} = \frac{d}{d x} [\frac{P_{1} (x)}{Q_{1} (x)}] + \frac{P_{2} (x)}{Q_{2} (x)}$ $degree P_{i} < degree Q_{i}$ $P (x) = 1$ $Q (x) = {(1 + 4 x + x^{2})}^{n}$ $Q^{'} (x) = 2 (2 + x) {(1 + 4 x + x^{2})}^{n - 1}$ $Q_{1} (x) = {(1 + 4 x + x^{2})}^{n - 1}$ $Q_{2} (x) = 1 + 4 x + x^{2}$ $\frac{1}{{(1 + 4 x + x^{2})}^{n}} = \frac{P_{1}^{'} (x) (1 + 4 x + x^{2}) - (n - 1) P_{1} (x) + P_{2} (x) {(1 + 4 x + x^{2})}^{n - 1}}{{(1 + 4 x + x^{2})}^{n}}$ $we cannot generally solve this I think$ $\int \frac{d x}{{(1 + 4 x + x^{2})}^{1}} = \frac{\sqrt{3}}{6} \ln ∣ \frac{x + 2 - \sqrt{3}}{x + 2 + \sqrt{3}} ∣ + C$ $\int \frac{d x}{{(1 + 4 x + x^{2})}^{2}} = - \frac{x + 2}{6 (1 + 4 x + x^{2})} + \frac{\sqrt{3}}{36} \ln ∣ \frac{x + 2 - \sqrt{3}}{x + 2 + \sqrt{3}} ∣ + C$ $\int \frac{d x}{{(1 + 4 x + x^{2})}^{3}} = \frac{x^{3} + 6 x^{2} + 7 x - 2}{24 {(1 + 4 x + x^{2})}^{2}} + \frac{\sqrt{3}}{144} \ln ∣ \frac{x + 2 - \sqrt{3}}{x + 2 + \sqrt{3}} ∣ + C$ $\int \frac{d x}{{(1 + 4 x + x^{2})}^{4}} = - \frac{5 x^{5} + 50 x^{4} + 160 x^{3} + 160 x^{2} + 19 x + 38}{432 {(1 + 4 x + x^{2})}^{3}} + \frac{5 \sqrt{3}}{2592} \ln ∣ \frac{x + 2 - \sqrt{3}}{x + 2 + \sqrt{3}} ∣ + C$ $. . .$ $we only need to find P_{1} and P_{2} and solve$ $the same integral \int \frac{d x}{1 + 4 x + x^{2}}$ $maybe someone can find a formula for the$ $coefficients . Sir Ramanujan, where have you$ $been ?; -)$
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