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Question Number 120230 by MehraGanesh last updated on 30/Oct/20

∫ (x^3 /((x +2)^4 )) dx OR ∫ ((cos x)/(1 + cos x)) dx a^→ = i^� − j^� + 3k^� and b^(→ ) = 2i^� − 7j^� + k^� .

$$\:\:\:\:\int\:\frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}\:+\mathrm{2}\right)^{\mathrm{4}} }\:\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$$\:\:\:\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx} \\ $$$$ \\ $$$$\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}\:=\:\hat {\mathrm{i}}\:−\:\hat {\mathrm{j}}\:+\:\mathrm{3}\hat {\mathrm{k}}\:\:\mathrm{and}\:\overset{\rightarrow\:} {\mathrm{b}}\:=\:\mathrm{2}\hat {\mathrm{i}}\:−\:\mathrm{7}\hat {\mathrm{j}}\:+\:\hat {\mathrm{k}}\:. \\ $$$$ \\ $$$$\:\:\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 30/Oct/20

∫((cosx)/(1+cosx))dx =∫1−(1/(1+cosx))dx =x−∫(1/(2cos^2 (x/2))) =x−(1/2)∫sec^2 (x/2)dx =x−tan(x/2)+C

$$\int\frac{{cosx}}{\mathrm{1}+{cosx}}{dx} \\ $$$$=\int\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{cosx}}{dx} \\ $$$$={x}−\int\frac{\mathrm{1}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$={x}−\frac{\mathrm{1}}{\mathrm{2}}\int{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx} \\ $$$$={x}−{tan}\frac{{x}}{\mathrm{2}}+{C} \\ $$

Answered by TANMAY PANACEA last updated on 30/Oct/20

∫(x^3 /((x+2)^4 ))dx t=x+2 ∫(((t−2)^3 )/t^4 )dt ∫((t^3 −6t^2 +12t−8)/t^4 )dt ∫(dt/t)−6∫t^(−2) dt+12∫t^(−3) dt−8∫t^(−4) dt =lnt+(6/t)−(6/t^2 )+(8/(3t^3 ))+c =ln(x+2)+(6/((x+2)))−(6/((x+2)^2 ))+(8/(3(x+2)^3 ))+c

$$\int\frac{{x}^{\mathrm{3}} }{\left({x}+\mathrm{2}\right)^{\mathrm{4}} }{dx} \\ $$$${t}={x}+\mathrm{2} \\ $$$$\int\frac{\left({t}−\mathrm{2}\right)^{\mathrm{3}} }{{t}^{\mathrm{4}} }{dt} \\ $$$$\int\frac{{t}^{\mathrm{3}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{12}{t}−\mathrm{8}}{{t}^{\mathrm{4}} }{dt} \\ $$$$\int\frac{{dt}}{{t}}−\mathrm{6}\int{t}^{−\mathrm{2}} {dt}+\mathrm{12}\int{t}^{−\mathrm{3}} {dt}−\mathrm{8}\int{t}^{−\mathrm{4}} {dt} \\ $$$$={lnt}+\frac{\mathrm{6}}{{t}}−\frac{\mathrm{6}}{{t}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}{t}^{\mathrm{3}} }+{c} \\ $$$$={ln}\left({x}+\mathrm{2}\right)+\frac{\mathrm{6}}{\left({x}+\mathrm{2}\right)}−\frac{\mathrm{6}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}\left({x}+\mathrm{2}\right)^{\mathrm{3}} }+{c} \\ $$

Answered by mathmax by abdo last updated on 30/Oct/20

I=∫ (x^3 /((x+2)^4 ))dx ⇒ I=_(x+2=t) ∫ (((t−2)^3 )/t^4 )dt =∫ ((t^3 −3t^2 .2 +3t.2^2 −2^3 )/t^4 )dt =∫ ((t^3 −6t^2 +12t−8)/t^4 )dt =∫ ((1/t)−(6/t^2 ) +((12)/t^3 )−(8/t^4 ))dt =ln∣t∣+(6/t)−(6/t^2 )+(8/(3t^3 )) +C =ln∣x+2∣+(6/(x+2))−(6/((x+2)^2 )) +(8/(3(x+2)^3 )) +C

$$\mathrm{I}=\int\:\frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{4}} }\mathrm{dx}\:\:\Rightarrow\:\mathrm{I}=_{\mathrm{x}+\mathrm{2}=\mathrm{t}} \:\:\:\int\:\:\frac{\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=\int\:\:\frac{\mathrm{t}^{\mathrm{3}} −\mathrm{3t}^{\mathrm{2}} .\mathrm{2}\:+\mathrm{3t}.\mathrm{2}^{\mathrm{2}} −\mathrm{2}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\int\:\:\frac{\mathrm{t}^{\mathrm{3}} −\mathrm{6t}^{\mathrm{2}} +\mathrm{12t}−\mathrm{8}}{\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=\int\:\:\left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{6}}{\mathrm{t}^{\mathrm{2}} }\:+\frac{\mathrm{12}}{\mathrm{t}^{\mathrm{3}} }−\frac{\mathrm{8}}{\mathrm{t}^{\mathrm{4}} }\right)\mathrm{dt}\:=\mathrm{ln}\mid\mathrm{t}\mid+\frac{\mathrm{6}}{\mathrm{t}}−\frac{\mathrm{6}}{\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3t}^{\mathrm{3}} }\:+\mathrm{C} \\ $$$$=\mathrm{ln}\mid\mathrm{x}+\mathrm{2}\mid+\frac{\mathrm{6}}{\mathrm{x}+\mathrm{2}}−\frac{\mathrm{6}}{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }\:+\frac{\mathrm{8}}{\mathrm{3}\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{3}} }\:+\mathrm{C} \\ $$

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