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Question Number 109509 by bemath last updated on 24/Aug/20

((bemath)/(Σ_(i=cooll) ^(nice) (joss)_i )) ∫ ((x^2 dx)/( (√(x^2 +25))))

$$\:\:\frac{{bemath}}{\underset{{i}={cooll}} {\overset{{nice}} {\sum}}\left({joss}\right)_{{i}} }\: \\ $$$$ \\ $$$$\int\:\frac{{x}^{\mathrm{2}} \:{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$

Answered by Dwaipayan Shikari last updated on 24/Aug/20

∫((x^2 +25)/( (√(x^2 +25))))−((25)/( (√(x^2 +25)))) ∫(√(x^2 +25))−25log(x+(√(x^2 +25))) (x^2 /2)(√(x^2 +25))+((25)/2)log(x+(√(x^2 +25)))−25log(x+(√(x^2 +25))) (x^2 /2)(√(x^2 +25)) −((25)/2)log(x+(√(x^2 +25)))+C

$$\int\frac{{x}^{\mathrm{2}} +\mathrm{25}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}−\frac{\mathrm{25}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$$$\int\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}−\mathrm{25}{log}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\right) \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}+\frac{\mathrm{25}}{\mathrm{2}}{log}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\right)−\mathrm{25}{log}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\right) \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\:−\frac{\mathrm{25}}{\mathrm{2}}{log}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\right)+{C} \\ $$

Answered by john santu last updated on 24/Aug/20

consider (d/dx) ((√(x^2 +25)) ) = ((2x)/(2(√(x^2 +25)))) = (x/( (√(x^2 +25)))) I= ∫ x ((x/( (√(x^2 +25)))) )dx = ∫ x d((√(x^2 +25))) by parts → { ((u=x⇒du=dx)),((v=(√(x^2 +25)))) :} I=x(√(x^2 +25)) −∫(√(x^2 +25)) dx the second integral set x = 5tan q ⇒dx = 5sec^2 q dq I_2 =∫(√(25(tan^2 q+1))) (5sec^2 q dq) I_2 =25∫sec^3 q dq I_2 =25((1/2)tan q.sec q +(1/2)ln ∣tan ((q/2)+(π/4))∣ +c I_2 = ((25)/2)((x/5).((√(x^2 +25))/5)) +((25)/2)ln ∣tan (((tan^(−1) ((x/5)))/2)+(π/4))∣+c

$${consider}\:\frac{{d}}{{dx}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\:\right)\:=\:\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$$${I}=\:\int\:{x}\:\left(\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}\:\right){dx}\:=\:\int\:{x}\:{d}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\right) \\ $$$${by}\:{parts}\:\rightarrow\begin{cases}{{u}={x}\Rightarrow{du}={dx}}\\{{v}=\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}\end{cases} \\ $$$${I}={x}\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\:−\int\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}\:{dx} \\ $$$${the}\:{second}\:{integral} \\ $$$${set}\:{x}\:=\:\mathrm{5tan}\:{q}\:\Rightarrow{dx}\:=\:\mathrm{5sec}\:^{\mathrm{2}} {q}\:{dq} \\ $$$${I}_{\mathrm{2}} =\int\sqrt{\mathrm{25}\left(\mathrm{tan}\:^{\mathrm{2}} {q}+\mathrm{1}\right)}\:\left(\mathrm{5sec}\:^{\mathrm{2}} {q}\:{dq}\right) \\ $$$${I}_{\mathrm{2}} =\mathrm{25}\int\mathrm{sec}\:^{\mathrm{3}} {q}\:{dq}\: \\ $$$${I}_{\mathrm{2}} =\mathrm{25}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:{q}.\mathrm{sec}\:{q}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{tan}\:\left(\frac{{q}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c}\right. \\ $$$${I}_{\mathrm{2}} =\:\frac{\mathrm{25}}{\mathrm{2}}\left(\frac{{x}}{\mathrm{5}}.\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}{\mathrm{5}}\right)\:+\frac{\mathrm{25}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{tan}\:\left(\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{5}}\right)}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid+{c} \\ $$

Answered by 1549442205PVT last updated on 25/Aug/20

Put x=5tant⇒dx=5(1+tan^2 t)dt ∫ ((x^2 dx)/( (√(x^2 +25))))=5∫((25tan^2 t(1+tan^2 t))/(5(√(1+tan^2 t))))dt =25∫((sin^2 t)/(cos^3 t))dt=25∫(((1−cos^2 t)/(cos^3 t))dt =25∫(dt/(cos^3 t))−25∫(dt/(cost))=25(A−B) B=∫sectdt=ln∣sect+tant∣ A=∫(dt/(cos^3 t))=∫((dtant)/(cost))=((tant)/(cost))−∫tant.((1/(cost)))′dt =((tant)/(cost))−∫tant.tant.sectdt =((tant)/(cost))−∫((sin^2 t)/(cos^3 t))=((tant)/(cost))−∫((1−cos^2 t)/(cos^3 t))dt =((tant)/(cost))+∫(dt/(cost))−A ⇒2A=((tant)/(cost))+ln∣sect+tant∣ ⇒A=(1/2).((tant)/(cost))+(1/2)ln∣sect+tant∣ Therefore,we obtain F=∫ ((x^2 dx)/( (√(x^2 +25))))=25(A−B) =((25)/2)(((tant)/(cost))+ln∣sect+tant∣)−25(ln∣sect+tant∣)+C =((25)/2)(((tant)/(cost))−ln∣sect+tant∣)+C since tant=(x/5),cost=(1/( (√(1+tan^2 t)))) =(1/( (√(1+(x^2 /(25))))))=(5/( (√(25+x^2 )))),sect=((√(25+x^2 ))/5) Consequently, F=((25)/2)(((x(√(25+x^2 )))/(25))−ln∣((√(25+x^2 ))/5)+(x/5)∣+C

$$\mathrm{Put}\:\mathrm{x}=\mathrm{5tant}\Rightarrow\mathrm{dx}=\mathrm{5}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt} \\ $$$$\int\:\frac{{x}^{\mathrm{2}} \:{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}=\mathrm{5}\int\frac{\mathrm{25tan}^{\mathrm{2}} \mathrm{t}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)}{\mathrm{5}\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}}}\mathrm{dt} \\ $$$$=\mathrm{25}\int\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}\mathrm{dt}=\mathrm{25}\int\left(\frac{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{t}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}\mathrm{dt}\right. \\ $$$$=\mathrm{25}\int\frac{\mathrm{dt}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}−\mathrm{25}\int\frac{\mathrm{dt}}{\mathrm{cost}}=\mathrm{25}\left(\mathrm{A}−\mathrm{B}\right) \\ $$$$\mathrm{B}=\int\mathrm{sectdt}=\mathrm{ln}\mid\mathrm{sect}+\mathrm{tant}\mid \\ $$$$\mathrm{A}=\int\frac{\mathrm{dt}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}=\int\frac{\mathrm{dtant}}{\mathrm{cost}}=\frac{\mathrm{tant}}{\mathrm{cost}}−\int\mathrm{tant}.\left(\frac{\mathrm{1}}{\mathrm{cost}}\right)'\mathrm{dt} \\ $$$$=\frac{\mathrm{tant}}{\mathrm{cost}}−\int\mathrm{tant}.\mathrm{tant}.\mathrm{sectdt} \\ $$$$=\frac{\mathrm{tant}}{\mathrm{cost}}−\int\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}=\frac{\mathrm{tant}}{\mathrm{cost}}−\int\frac{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{t}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{tant}}{\mathrm{cost}}+\int\frac{\mathrm{dt}}{\mathrm{cost}}−\mathrm{A} \\ $$$$\Rightarrow\mathrm{2A}=\frac{\mathrm{tant}}{\mathrm{cost}}+\mathrm{ln}\mid\mathrm{sect}+\mathrm{tant}\mid \\ $$$$\Rightarrow\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{tant}}{\mathrm{cost}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{sect}+\mathrm{tant}\mid \\ $$$$\mathrm{Therefore},\mathrm{we}\:\mathrm{obtain} \\ $$$$\mathrm{F}=\int\:\frac{{x}^{\mathrm{2}} \:{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}=\mathrm{25}\left(\mathrm{A}−\mathrm{B}\right) \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}\left(\frac{\mathrm{tant}}{\mathrm{cost}}+\mathrm{ln}\mid\mathrm{sect}+\mathrm{tant}\mid\right)−\mathrm{25}\left(\mathrm{ln}\mid\mathrm{sect}+\mathrm{tant}\mid\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}\left(\frac{\boldsymbol{\mathrm{tant}}}{\boldsymbol{\mathrm{cost}}}−\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{sect}}+\boldsymbol{\mathrm{tant}}\mid\right)+\boldsymbol{\mathrm{C}}\: \\ $$$$\mathrm{since}\:\mathrm{tant}=\frac{\mathrm{x}}{\mathrm{5}},\mathrm{cost}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{25}}}}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{25}+\mathrm{x}^{\mathrm{2}} }},\mathrm{sect}=\frac{\sqrt{\mathrm{25}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{5}} \\ $$$$\mathrm{Consequently}, \\ $$$$\boldsymbol{\mathrm{F}}=\frac{\mathrm{25}}{\mathrm{2}}\left(\frac{\boldsymbol{\mathrm{x}}\sqrt{\mathrm{25}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}{\mathrm{25}}−\boldsymbol{\mathrm{ln}}\mid\frac{\sqrt{\mathrm{25}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}{\mathrm{5}}+\frac{\boldsymbol{\mathrm{x}}}{\mathrm{5}}\mid+\boldsymbol{\mathrm{C}}\right. \\ $$

Answered by mathmax by abdo last updated on 24/Aug/20

I =∫ (x^2 /(√(x^2 +25))) dx by parts u^′ =(x/(√(x^2 +25))) snd v =x ⇒ I =x(√(x^2 +25)) −∫ (√(x^2 +25)) dx but ∫ (√(x^2 +25))dx =_(x=5sh(t)) ∫ 5ch(t)(5cht)dt =25 ∫ ((1+ch(2t))/2)dt =((25t)/2) +((25)/4)sh(2t) +c =((25t)/2) +((25)/2)sh(t)ch(t) +c =((25)/2) argsh((x/5))++(5/2)x(√(1+((x/5))^2 )) +c =((25)/2)ln((x/5)+(√(1+(x^2 /(25))))) +((5x)/2)(√(1+(x^2 /(25)))) +c ⇒ I =x(√(x^2 +25))−((25)/2)ln((x/5) +(√(1+(x^2 /(25)))))+((5x)/2)(√(1+(x^2 /(25)))) +C

$$\mathrm{I}\:=\int\:\frac{\mathrm{x}^{\mathrm{2}} }{\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{25}}}\:\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{u}^{'} \:=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{25}}}\:\mathrm{snd}\:\mathrm{v}\:=\mathrm{x}\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{25}}\:−\int\:\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{25}}\:\mathrm{dx}\:\:\mathrm{but}\: \\ $$$$\int\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{25}}\mathrm{dx}\:=_{\mathrm{x}=\mathrm{5sh}\left(\mathrm{t}\right)} \:\:\:\int\:\:\mathrm{5ch}\left(\mathrm{t}\right)\left(\mathrm{5cht}\right)\mathrm{dt}\:=\mathrm{25}\:\int\:\frac{\mathrm{1}+\mathrm{ch}\left(\mathrm{2t}\right)}{\mathrm{2}}\mathrm{dt} \\ $$$$=\frac{\mathrm{25t}}{\mathrm{2}}\:\:+\frac{\mathrm{25}}{\mathrm{4}}\mathrm{sh}\left(\mathrm{2t}\right)\:+\mathrm{c}\:=\frac{\mathrm{25t}}{\mathrm{2}}\:+\frac{\mathrm{25}}{\mathrm{2}}\mathrm{sh}\left(\mathrm{t}\right)\mathrm{ch}\left(\mathrm{t}\right)\:+\mathrm{c} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}\:\mathrm{argsh}\left(\frac{\mathrm{x}}{\mathrm{5}}\right)++\frac{\mathrm{5}}{\mathrm{2}}\mathrm{x}\sqrt{\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{5}}\right)^{\mathrm{2}} }\:\:+\mathrm{c} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{5}}+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{25}}}\right)\:+\frac{\mathrm{5x}}{\mathrm{2}}\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{25}}}\:+\mathrm{c}\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{25}}−\frac{\mathrm{25}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{5}}\:+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{25}}}\right)+\frac{\mathrm{5x}}{\mathrm{2}}\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{25}}}\:+\mathrm{C} \\ $$

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