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Question Number 82283 by M±th+et£s last updated on 19/Feb/20

∫x^3 (√(x^3 +1)) dx

$$\int{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dx} \\ $$

Commented by MJS last updated on 20/Feb/20

it′s impossible for me to solve this

$$\mathrm{it}'\mathrm{s}\:\mathrm{impossible}\:\mathrm{for}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this} \\ $$

Commented by M±th+et£s last updated on 20/Feb/20

i cant solve it to sir i posted it asking for help but i think that its special integral

$${i}\:{cant}\:{solve}\:{it}\:{to}\:{sir}\: \\ $$$${i}\:{posted}\:{it}\:{asking}\:{for}\:{help}\:\: \\ $$$${but}\:{i}\:{think}\:{that}\:{its}\:{special}\:{integral} \\ $$

Answered by mind is power last updated on 20/Feb/20

Hello (√(1+r))=1+(r/2)+Σ_(k≥2) (−1)^(k−1) .(((2k−3)!!)/(2^k k!)).x^k =1+Σ_(k≥1) (−1)^(k−1) (((2k−2)!)/(2^(2k−1) .k((k−1)!)^2 ))x^k ∫x^3 (√(1+x^3 ))dx =∫x^3 (1+Σ_(k≥1) (−1)^(k−1) (((2k−2)!)/(2^(2k−1) k!.(k−1)!)).x^(3k) )dx =(x^4 /4)+Σ_(k≥1) (((−1)^(k−1) (2k−2)!4)/(2^(2k−1) (k−1)!.(3k+4).k!))x^(3k+4) =(x^4 /4)(1+Σ_(k≥1) (((−1)^(k−1) (2k−2)!)/(2^(2k−1) (k−1)!(3k+4))).(x^(3k) /(k!))) =(x^4 /4)(1+Σ_(k≥1) (((−1)^(k−1) .4.(2k−3)!!)/(2^k (k+4))).(x^(3k) /(k!))) =(x^4 /4)(1+Σ_(k≥1) (((−(1/2)).......(−(1/2)+k−1)(4/3).....(k+(4/3)−1))/(((7/3))......((7/3)+k−1))) .((x^3 /2))^k .(1/(k!)) =(x^4 /4) _2 F_1 (−(1/2),(4/3);(7/3);((x^3 /2)))+c

$${Hello}\: \\ $$$$\sqrt{\mathrm{1}+{r}}=\mathrm{1}+\frac{{r}}{\mathrm{2}}+\underset{{k}\geqslant\mathrm{2}} {\sum}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} .\frac{\left(\mathrm{2}{k}−\mathrm{3}\right)!!}{\mathrm{2}^{{k}} {k}!}.{x}^{{k}} \\ $$$$=\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \frac{\left(\mathrm{2}{k}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} .{k}\left(\left({k}−\mathrm{1}\right)!\right)^{\mathrm{2}} }{x}^{{k}} \\ $$$$\int{x}^{\mathrm{3}} \sqrt{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$$$=\int{x}^{\mathrm{3}} \left(\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \frac{\left(\mathrm{2}{k}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} {k}!.\left({k}−\mathrm{1}\right)!}.{x}^{\mathrm{3}{k}} \right){dx} \\ $$$$=\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left(\mathrm{2}{k}−\mathrm{2}\right)!\mathrm{4}}{\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!.\left(\mathrm{3}{k}+\mathrm{4}\right).{k}!}{x}^{\mathrm{3}{k}+\mathrm{4}} \\ $$$$=\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\left(\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left(\mathrm{2}{k}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left(\mathrm{3}{k}+\mathrm{4}\right)}.\frac{{x}^{\mathrm{3}{k}} }{{k}!}\right) \\ $$$$=\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\left(\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} .\mathrm{4}.\left(\mathrm{2}{k}−\mathrm{3}\right)!!}{\mathrm{2}^{{k}} \left({k}+\mathrm{4}\right)}.\frac{{x}^{\mathrm{3}{k}} }{{k}!}\right) \\ $$$$=\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\left(\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right).......\left(−\frac{\mathrm{1}}{\mathrm{2}}+{k}−\mathrm{1}\right)\frac{\mathrm{4}}{\mathrm{3}}.....\left({k}+\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}\right)}{\left(\frac{\mathrm{7}}{\mathrm{3}}\right)......\left(\frac{\mathrm{7}}{\mathrm{3}}+{k}−\mathrm{1}\right)}\:.\left(\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!}\right. \\ $$$$=\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{4}}{\mathrm{3}};\frac{\mathrm{7}}{\mathrm{3}};\left(\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\right)\right)+{c} \\ $$

Commented by MJS last updated on 20/Feb/20

I thought there was a possibility using series but I have never learned these things...

$$\mathrm{I}\:\mathrm{thought}\:\mathrm{there}\:\mathrm{was}\:\mathrm{a}\:\mathrm{possibility}\:\mathrm{using}\:\mathrm{series} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{have}\:\mathrm{never}\:\mathrm{learned}\:\mathrm{these}\:\mathrm{things}... \\ $$

Commented by M±th+et£s last updated on 20/Feb/20

great sir thank you so much . god bless you

$${great}\:{sir}\:{thank}\:{you}\:{so}\:{much}\:.\:{god}\:{bless}\:{you} \\ $$

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