1-4x-x-2-m-dx-

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

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)}^{\boldsymbol m - \boldsymbol k} \boldsymbol 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 \dots

Answered by Hope last updated on 10/Jul/19

\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 \dots 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

\dots

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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