Question Number 63844 by mmkkmm000m last updated on 10/Jul/19
$$\int\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{m}} {dx} \\ $$
Commented by mathmax by abdo last updated on 10/Jul/19
$${let}\:{A}_{{m}} =\int\:\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{{m}} \:{dx}\:\Rightarrow \\ $$$${A}_{{m}} =\int\:\left({x}^{\mathrm{2}} +\mathrm{4}{x}\:+\mathrm{4}−\mathrm{3}\right)^{{m}} {dx}\:=\int\left\{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\right\}^{{m}} {dx} \\ $$$$=\int\:\:\sum_{{k}=\mathrm{0}} ^{{m}} {C}_{{k}} ^{{m}} \:\:\left({x}+\mathrm{2}\right)^{\mathrm{2}{k}} \:\left(−\mathrm{3}\right)^{\boldsymbol{{m}}−\boldsymbol{{k}}} \boldsymbol{{dx}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{m}} \:\left(−\mathrm{3}\right)^{{m}−{k}} \:\:{C}_{{m}} ^{{k}} \:\int\:\left({x}+\mathrm{2}\right)^{\mathrm{2}{k}} \:{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{m}} \:\left(−\mathrm{3}\right)^{{m}−{k}} \:{C}_{{m}} ^{{k}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\left({x}+\mathrm{2}\right)^{\mathrm{2}{k}+\mathrm{1}} \:+{C} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{m}} \:\left(−\mathrm{3}\right)^{{m}−{k}} \:\frac{{C}_{{m}} ^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\left({x}+\mathrm{2}\right)^{\mathrm{2}{k}+\mathrm{1}} \:+{C}\:. \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 10/Jul/19
$${at}\:{line}\:\mathrm{3}\:{change}\:{C}_{{k}} ^{{m}} \:{by}\:{C}_{{m}} ^{{k}} \\ $$
Commented by mathmax by abdo last updated on 10/Jul/19
$${i}\:{have}\:{taken}\:{m}\:{from}\:{N}. \\ $$
Answered by MJS last updated on 10/Jul/19
$$\left({a}+{b}+{c}\right)^{{n}} \:\mathrm{with}\:{n}\in\mathbb{N} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:{trinomial}\:{coefficients} \\ $$$$\left({a}+{b}+{c}\right)^{{n}} =\underset{{i},\:{j},\:{k}} {\sum}\begin{pmatrix}{{n}}\\{{i},\:{j},\:{k}}\end{pmatrix}{a}^{{i}} {b}^{{j}} {c}^{{k}} \:\mathrm{with}\:{i}+{j}+{k}={n} \\ $$$$\mathrm{and}\:\begin{pmatrix}{{n}}\\{{i},\:{j},\:{k}}\end{pmatrix}=\frac{{n}!}{{i}!{j}!{k}!} \\ $$$$ \\ $$$$\mathrm{for}\:{m},\:{i},\:{j},\:{k}\in\mathbb{N}\wedge{i}+{j}+{k}={m}\:\mathrm{we}\:\mathrm{get} \\ $$$$\int\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{m}} {dx}= \\ $$$$=\int\underset{{i},\:{j},\:{k}} {\sum}\begin{pmatrix}{{m}}\\{{i},\:{j},\:{k}}\end{pmatrix}\mathrm{1}^{{i}} \left(\mathrm{4}{x}\right)^{{j}} \left({x}^{\mathrm{2}} \right)^{{k}} {dx}= \\ $$$$=\underset{{i},\:{j},\:{k}} {\sum}\left(\frac{{m}!}{{i}!{j}!{k}!}\int\mathrm{4}^{{j}} {x}^{{j}+\mathrm{2}{k}} {dx}\right)= \\ $$$$=\underset{{i},\:{j},\:{k}} {\sum}\left(\frac{\mathrm{4}^{{j}} {m}!}{\left({j}+\mathrm{2}{k}+\mathrm{1}\right){i}!{j}!{k}!}{x}^{{j}+\mathrm{2}{k}+\mathrm{1}} \right)+{C} \\ $$$$ \\ $$$$\mathrm{for}\:{m}\in\mathbb{Z}^{−} \:\mathrm{and}\:{m}\in\mathbb{Q}\:\mathrm{I}'\mathrm{m}\:\mathrm{still}\:\mathrm{trying}… \\ $$
Answered by Hope last updated on 10/Jul/19
$$\int\left\{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\right\}^{{m}} {dx} \\ $$$${y}=\left({x}+\mathrm{2}\right)\:\:\:\mathrm{3}={a}^{\mathrm{2}} \\ $$$${I}_{{m}} =\int\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}} {dy} \\ $$$${I}_{{m}} =\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}} {y}−{m}\int\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}−\mathrm{1}} ×\mathrm{2}{y}×{ydy} \\ $$$$=\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}} ×{y}−\mathrm{2}{m}\int\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}−\mathrm{1}} \left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} +{a}^{\mathrm{2}} \right){dy} \\ $$$$=\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}} ×{y}−\mathrm{2}{mI}_{{m}} −\mathrm{2}{ma}^{\mathrm{2}} {I}_{{m}−\mathrm{1}} \\ $$$${I}_{{m}} \left(\mathrm{1}+\mathrm{2}{m}\right)={y}\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}} −\mathrm{2}{ma}^{\mathrm{2}} {I}_{{m}−\mathrm{1}} \\ $$$${I}_{{m}} =\frac{{y}\left({y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{{m}} }{\mathrm{1}+\mathrm{2}{m}}−\frac{\mathrm{2}{ma}^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{m}}{I}_{{m}−\mathrm{1}} \\ $$$$\int\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{m}} {dx} \\ $$$$=\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{m}} }{\mathrm{1}+\mathrm{2}{m}}−\frac{\mathrm{2}{m}×\mathrm{3}}{\mathrm{1}+\mathrm{2}{m}}\int\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{m}−\mathrm{1}} {dx} \\ $$$${pls}\:{check}…{Tanmay} \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 10/Jul/19
![∫(1+4x+x^2 )^m dx with m∈Z^− ∫(dx/((1+4x+x^2 )^n )) with n∈N trying Ostrogradski′s Method ∫((P(x))/(Q(x)))dx=((P_1 (x))/(Q_1 (x)))+∫((P_2 (x))/(Q_2 (x)))dx Q_1 (x)=gcd (Q(x), Q′(x)) Q_2 (x)=((Q(x))/(Q_1 (x))) P_1 (x), P_2 (x) can be found by comparing the constant factors of ((P(x))/(Q(x)))=(d/dx)[((P_1 (x))/(Q_1 (x)))]+((P_2 (x))/(Q_2 (x))) degree P_i <degree Q_i P(x)=1 Q(x)=(1+4x+x^2 )^n Q′(x)=2(2+x)(1+4x+x^2 )^(n−1) Q_1 (x)=(1+4x+x^2 )^(n−1) Q_2 (x)=1+4x+x^2 (1/((1+4x+x^2 )^n ))=((P_1 ′(x)(1+4x+x^2 )−(n−1)P_1 (x)+P_2 (x)(1+4x+x^2 )^(n−1) )/((1+4x+x^2 )^n )) we cannot generally solve this I think ∫(dx/((1+4x+x^2 )^1 ))=((√3)/6)ln ∣((x+2−(√3))/(x+2+(√3)))∣ +C ∫(dx/((1+4x+x^2 )^2 ))=−((x+2)/(6(1+4x+x^2 )))+((√3)/(36))ln ∣((x+2−(√3))/(x+2+(√3)))∣ +C ∫(dx/((1+4x+x^2 )^3 ))=((x^3 +6x^2 +7x−2)/(24(1+4x+x^2 )^2 ))+((√3)/(144))ln ∣((x+2−(√3))/(x+2+(√3)))∣ +C ∫(dx/((1+4x+x^2 )^4 ))=−((5x^5 +50x^4 +160x^3 +160x^2 +19x+38)/(432(1+4x+x^2 )^3 ))+((5(√3))/(2592))ln ∣((x+2−(√3))/(x+2+(√3)))∣ +C ... we only need to find P_1 and P_2 and solve the same integral ∫(dx/(1+4x+x^2 )) maybe someone can find a formula for the coefficients. Sir Ramanujan, where have you been? ;−)](https://www.tinkutara.com/question/Q63863.png)
$$\int\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{m}} {dx}\:\mathrm{with}\:{m}\in\mathbb{Z}^{−} \\ $$$$\int\frac{{dx}}{\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{n}} }\:\mathrm{with}\:{n}\in\mathbb{N} \\ $$$$\mathrm{trying}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method} \\ $$$$\int\frac{{P}\left({x}\right)}{{Q}\left({x}\right)}{dx}=\frac{{P}_{\mathrm{1}} \left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)}+\int\frac{{P}_{\mathrm{2}} \left({x}\right)}{{Q}_{\mathrm{2}} \left({x}\right)}{dx} \\ $$$${Q}_{\mathrm{1}} \left({x}\right)=\mathrm{gcd}\:\left({Q}\left({x}\right),\:{Q}'\left({x}\right)\right) \\ $$$${Q}_{\mathrm{2}} \left({x}\right)=\frac{{Q}\left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)} \\ $$$${P}_{\mathrm{1}} \left({x}\right),\:{P}_{\mathrm{2}} \left({x}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{by}\:\mathrm{comparing}\:\mathrm{the} \\ $$$$\mathrm{constant}\:\mathrm{factors}\:\mathrm{of} \\ $$$$\frac{{P}\left({x}\right)}{{Q}\left({x}\right)}=\frac{{d}}{{dx}}\left[\frac{{P}_{\mathrm{1}} \left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)}\right]+\frac{{P}_{\mathrm{2}} \left({x}\right)}{{Q}_{\mathrm{2}} \left({x}\right)} \\ $$$$\mathrm{degree}\:{P}_{{i}} \:<\mathrm{degree}\:{Q}_{{i}} \\ $$$$ \\ $$$${P}\left({x}\right)=\mathrm{1} \\ $$$${Q}\left({x}\right)=\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{n}} \\ $$$${Q}'\left({x}\right)=\mathrm{2}\left(\mathrm{2}+{x}\right)\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \\ $$$${Q}_{\mathrm{1}} \left({x}\right)=\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \\ $$$${Q}_{\mathrm{2}} \left({x}\right)=\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{n}} }=\frac{{P}_{\mathrm{1}} '\left({x}\right)\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)−\left({n}−\mathrm{1}\right){P}_{\mathrm{1}} \left({x}\right)+{P}_{\mathrm{2}} \left({x}\right)\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{n}} } \\ $$$$\mathrm{we}\:\mathrm{cannot}\:\mathrm{generally}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{I}\:\mathrm{think} \\ $$$$\int\frac{{dx}}{\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{1}} }=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{2}−\sqrt{\mathrm{3}}}{{x}+\mathrm{2}+\sqrt{\mathrm{3}}}\mid\:+{C} \\ $$$$\int\frac{{dx}}{\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=−\frac{{x}+\mathrm{2}}{\mathrm{6}\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)}+\frac{\sqrt{\mathrm{3}}}{\mathrm{36}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{2}−\sqrt{\mathrm{3}}}{{x}+\mathrm{2}+\sqrt{\mathrm{3}}}\mid\:+{C} \\ $$$$\int\frac{{dx}}{\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}}{\mathrm{24}\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\sqrt{\mathrm{3}}}{\mathrm{144}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{2}−\sqrt{\mathrm{3}}}{{x}+\mathrm{2}+\sqrt{\mathrm{3}}}\mid\:+{C} \\ $$$$\int\frac{{dx}}{\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }=−\frac{\mathrm{5}{x}^{\mathrm{5}} +\mathrm{50}{x}^{\mathrm{4}} +\mathrm{160}{x}^{\mathrm{3}} +\mathrm{160}{x}^{\mathrm{2}} +\mathrm{19}{x}+\mathrm{38}}{\mathrm{432}\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }+\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2592}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{2}−\sqrt{\mathrm{3}}}{{x}+\mathrm{2}+\sqrt{\mathrm{3}}}\mid\:+{C} \\ $$$$… \\ $$$$\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:{P}_{\mathrm{1}} \:\mathrm{and}\:{P}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{solve} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{integral}\:\int\frac{{dx}}{\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{maybe}\:\mathrm{someone}\:\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{coefficients}.\:\mathrm{Sir}\:\mathrm{Ramanujan},\:\mathrm{where}\:\mathrm{have}\:\mathrm{you} \\ $$$$\left.\mathrm{been}?\:;−\right) \\ $$