x-3-x-4-dx-

Question Number 10880 by Saham last updated on 28/Feb/17

$$\left.\int\:\left(\mathrm{x}\:+\:\mathrm{3}\right)\sqrt{\left(\mathrm{x}\:+\:\mathrm{4}\right.}\right)\:\mathrm{dx}\: \\ $$

Answered by geovane10math last updated on 28/Feb/17

∫(x + 3)(√(x + 4)) dx = ∫x(√(x + 4)) dx + ∫3(√(x + 4)) dx First integral: ∫x(√(x + 4)) dx = x + 4 = u ⇒ (du/dx) = 1 ⇒ du = dx ∫(u − 4)(√u) du = ∫u(√u) − 4(√u) du = = ∫u(√u) du − ∫4(√u) du = ∫u^(3/2) du − 4∫u^(1/2) du = (u^(5/2) /(5/2)) − 4∙(u^(3/2) /(3/2)) = ((2u^(5/2) )/5) − ((8u^(3/2) )/3) x + 4 = u, so = ((2(x + 4)^(5/2) )/5) − ((8(x + 4)^(3/2) )/3) First integral: ∫x(√(x+4)) dx = ((2(x + 4)^(5/2) )/5) − ((8(x + 4)^(3/2) )/3) + c_0 Second integral: ∫3(√(x + 4)) dx = 3∫(√(x + 4)) dx x + 4 = u ⇒ (du/dx) = 1 ⇒ du = dx = 3∫(√u) du = 3∫u^(1/2) du = 3∙(u^(3/2) /(3/2)) = ((6u^(3/2) )/2) = = 3u^(3/2) = 3(x + 4)^(3/2) Second integral: ∫3(√(x + 4)) dx = 3(x + 4)^(3/2) + c_1 So, ∫(x + 3)(√(x + 4)) dx = ((2(x + 4)^(5/2) )/5) − ((8(x + 4)^(3/2) )/3) + 3(x + 4)^(3/2) + C

$$\int\left({x}\:+\:\mathrm{3}\right)\sqrt{{x}\:+\:\mathrm{4}}\:\mathrm{dx}\:=\:\int{x}\sqrt{{x}\:+\:\mathrm{4}}\:\mathrm{dx}\:+\:\int\mathrm{3}\sqrt{{x}\:+\:\mathrm{4}}\:\mathrm{dx} \\ $$$$\mathrm{First}\:\mathrm{integral}: \\ $$$$\int{x}\sqrt{{x}\:+\:\mathrm{4}}\:\mathrm{dx}\:= \\ $$$${x}\:+\:\mathrm{4}\:=\:{u}\:\Rightarrow\:\frac{{du}}{{dx}}\:=\:\mathrm{1}\:\Rightarrow\:{du}\:=\:{dx} \\ $$$$\int\left({u}\:−\:\mathrm{4}\right)\sqrt{{u}}\:\mathrm{du}\:=\:\int{u}\sqrt{{u}}\:−\:\mathrm{4}\sqrt{{u}}\:\mathrm{du}\:=\: \\ $$$$=\:\int{u}\sqrt{{u}}\:\mathrm{du}\:−\:\int\mathrm{4}\sqrt{{u}}\:\mathrm{du}\:=\:\int{u}^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{du}\:−\:\mathrm{4}\int{u}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{du} \\ $$$$=\:\frac{{u}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\frac{\mathrm{5}}{\mathrm{2}}}\:−\:\mathrm{4}\centerdot\frac{{u}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}\:=\:\frac{\mathrm{2}{u}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}\:−\:\frac{\mathrm{8}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}\: \\ $$$${x}\:+\:\mathrm{4}\:=\:{u},\:\mathrm{so} \\ $$$$\:=\:\frac{\mathrm{2}\left({x}\:+\:\mathrm{4}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}\:−\:\frac{\mathrm{8}\left({x}\:+\:\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}} \\ $$$$\mathrm{First}\:\mathrm{integral}: \\ $$$$\int\boldsymbol{{x}}\sqrt{\boldsymbol{{x}}+\mathrm{4}}\:\boldsymbol{\mathrm{dx}}\:=\:\frac{\mathrm{2}\left(\boldsymbol{{x}}\:+\:\mathrm{4}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}\:−\:\frac{\mathrm{8}\left(\boldsymbol{{x}}\:+\:\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}\:+\:\boldsymbol{{c}}_{\mathrm{0}} \\ $$$$\: \\ $$$$\mathrm{Second}\:\mathrm{integral}: \\ $$$$\int\mathrm{3}\sqrt{{x}\:+\:\mathrm{4}}\:\mathrm{dx}\:=\:\mathrm{3}\int\sqrt{{x}\:+\:\mathrm{4}}\:\mathrm{dx} \\ $$$${x}\:+\:\mathrm{4}\:=\:{u}\:\Rightarrow\:\frac{{du}}{{dx}}\:=\:\mathrm{1}\:\Rightarrow\:{du}\:=\:{dx} \\ $$$$\:=\:\mathrm{3}\int\sqrt{{u}}\:\mathrm{du}\:=\:\mathrm{3}\int{u}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{du}\:=\:\mathrm{3}\centerdot\frac{{u}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}\:=\:\frac{\mathrm{6}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}}\:=\:\: \\ $$$$\:=\:\mathrm{3}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\:\mathrm{3}\left({x}\:+\:\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\mathrm{Second}\:\mathrm{integral}: \\ $$$$\int\mathrm{3}\sqrt{\boldsymbol{{x}}\:+\:\mathrm{4}}\:\boldsymbol{\mathrm{dx}}\:=\:\mathrm{3}\left(\boldsymbol{{x}}\:+\:\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\:\boldsymbol{{c}}_{\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{So}, \\ $$$$\int\left({x}\:+\:\mathrm{3}\right)\sqrt{{x}\:+\:\mathrm{4}}\:\mathrm{dx}\:=\:\frac{\mathrm{2}\left({x}\:+\:\mathrm{4}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}\:−\:\frac{\mathrm{8}\left({x}\:+\:\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}\:+\:\mathrm{3}\left({x}\:+\:\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\:{C}\: \\ $$

Commented by Saham last updated on 28/Feb/17

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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