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Question Number 145671 by mathlove last updated on 07/Jul/21
Commented by mathlove last updated on 07/Jul/21
pleas help me
$${pleas}\:{help}\:{me} \\ $$
Answered by imjagoll last updated on 07/Jul/21
(11) ∫ ((√(x^2 −9))/((x^2 +3x)(x−3))) dx = ∫ (((√(x+3)) (√(x−3)))/(x((√(x+3)))^2 ((√(x−3)))^2 )) dx = ∫ (dx/(x (√(x^2 −9)))) let x = 3sec t ⇒dx=3sec t tan t dt I=∫ ((3sec t tan t dt)/(3sec t (√(9tan^2 t)))) I=∫(1/3) dt = (1/3)arcsec ((x/3))+c
$$\left(\mathrm{11}\right)\:\int\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{9}}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}\right)\left(\mathrm{x}−\mathrm{3}\right)}\:\mathrm{dx}\: \\ $$$$\:=\:\int\:\frac{\sqrt{\mathrm{x}+\mathrm{3}}\:\sqrt{\mathrm{x}−\mathrm{3}}}{\mathrm{x}\left(\sqrt{\mathrm{x}+\mathrm{3}}\right)^{\mathrm{2}} \:\left(\sqrt{\mathrm{x}−\mathrm{3}}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{9}}}\: \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\mathrm{3sec}\:\mathrm{t}\:\Rightarrow\mathrm{dx}=\mathrm{3sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{3sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:\mathrm{dt}}{\mathrm{3sec}\:\mathrm{t}\:\sqrt{\mathrm{9tan}\:^{\mathrm{2}} \mathrm{t}}} \\ $$$$\mathrm{I}=\int\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsec}\:\left(\frac{\mathrm{x}}{\mathrm{3}}\right)+\mathrm{c}\: \\ $$
Commented by mathlove last updated on 07/Jul/21
and ohter
$${and}\:{ohter} \\ $$
Answered by puissant last updated on 07/Jul/21
16) t=x^2 ⇒ dt=2xdx ⇒ dx=(dt/(2x)) I=∫(x^3 /((x^2 +1)^3 ))×(dt/(2x)) = (1/2)∫(x^2 /((x^2 +1)^3 ))dt =(1/2)∫(t/((t+1)^3 ))dt=(1/2)∫(1/((t+1)^2 ))dt−(1/2)∫(1/((t+1)^3 ))dt ⇒I=−(1/(2(t+1)))+(1/(4(t+1)^2 ))+C I=(1/(4(x^2 +1)^2 ))−(1/(2(x^2 +1)))+C..
$$\left.\mathrm{16}\right) \\ $$$$\mathrm{t}=\mathrm{x}^{\mathrm{2}} \Rightarrow\:\mathrm{dt}=\mathrm{2xdx}\:\Rightarrow\:\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{2x}} \\ $$$$\mathrm{I}=\int\frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }×\frac{\mathrm{dt}}{\mathrm{2x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{t}}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dt} \\ $$$$\Rightarrow\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{t}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{C} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}+\mathrm{C}.. \\ $$
Commented by mathlove last updated on 07/Jul/21
and other
$${and}\:{other} \\ $$
Answered by iloveisrael last updated on 07/Jul/21
(9) ∫ ((cos 3u)/( (4+4sin 3u+sin^2 3u)^2 )) du =∫ ((cos 3u)/((2+sin 3u)^2 )) du =(1/3)∫ ((d(2+sin 3u))/((2+sin 3u)^2 )) =−(1/(3(2+sin 3u))) + c
$$\left(\mathrm{9}\right)\:\int\:\frac{\mathrm{cos}\:\mathrm{3}{u}}{\:\left(\mathrm{4}+\mathrm{4sin}\:\mathrm{3}{u}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}{u}\right)^{\mathrm{2}} }\:{du} \\ $$$$=\int\:\frac{\mathrm{cos}\:\mathrm{3}{u}}{\left(\mathrm{2}+\mathrm{sin}\:\mathrm{3}{u}\right)^{\mathrm{2}} }\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{{d}\left(\mathrm{2}+\mathrm{sin}\:\mathrm{3}{u}\right)}{\left(\mathrm{2}+\mathrm{sin}\:\mathrm{3}{u}\right)^{\mathrm{2}} }\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{2}+\mathrm{sin}\:\mathrm{3}{u}\right)}\:+\:{c}\: \\ $$
Commented by mathlove last updated on 08/Jul/21
thanks sir and other
$${thanks}\:{sir}\:\:{and}\:{other} \\ $$
Answered by bobhans last updated on 07/Jul/21
(5)∫ ((e^x +e^(2x) )/( (√(e^x +1)))) dx = ∫ ((e^x (e^x +1))/( (√(e^x +1)))) dx =∫e^x (√(1+e^x )) dx = ∫(√(1+e^x )) d(1+e^x ) = (2/3) (√((1+e^x )^3 )) + c
$$\left(\mathrm{5}\right)\int\:\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{\mathrm{2x}} }{\:\sqrt{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{e}^{\mathrm{x}} \left(\mathrm{e}^{\mathrm{x}} +\mathrm{1}\right)}{\:\sqrt{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}}\:\mathrm{dx} \\ $$$$=\int\mathrm{e}^{\mathrm{x}} \:\sqrt{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\:\mathrm{dx}\:=\:\int\sqrt{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\:\mathrm{d}\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right) \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:\sqrt{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)^{\mathrm{3}} }\:+\:\mathrm{c}\: \\ $$
Answered by bobhans last updated on 07/Jul/21
(15)∫ (((3+x)^2 −6x)/( (√((9+x^2 )^5 )))) dx = ∫ ((x^2 +9)/( (√((x^2 +9)^5 )))) dx = ∫ (dx/( (√((x^2 +9)^3 )))) ; x=3tan u ; sin u=(x/( (√(9+x^2 )))) =∫ ((3sec^2 u)/( (√((9sec^2 u)^3 )))) = (1/9)∫ (du/(sec u)) =(1/9)∫ cos u du = (1/9)sin u+c = (x/(9(√(9+x^2 )))) + c
$$\left(\mathrm{15}\right)\int\:\frac{\left(\mathrm{3}+\mathrm{x}\right)^{\mathrm{2}} −\mathrm{6x}}{\:\sqrt{\left(\mathrm{9}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{5}} }}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{9}}{\:\sqrt{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{5}} }}\:\mathrm{dx} \\ $$$$=\:\int\:\frac{\mathrm{dx}}{\:\sqrt{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{3}} }}\:;\:\mathrm{x}=\mathrm{3tan}\:\mathrm{u}\:;\:\mathrm{sin}\:\mathrm{u}=\frac{\mathrm{x}}{\:\sqrt{\mathrm{9}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$=\int\:\frac{\mathrm{3sec}\:^{\mathrm{2}} \mathrm{u}}{\:\sqrt{\left(\mathrm{9sec}\:^{\mathrm{2}} \mathrm{u}\right)^{\mathrm{3}} }}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\int\:\frac{\mathrm{du}}{\mathrm{sec}\:\mathrm{u}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\int\:\mathrm{cos}\:\mathrm{u}\:\mathrm{du}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\mathrm{sin}\:\mathrm{u}+\mathrm{c}\:=\:\frac{\mathrm{x}}{\mathrm{9}\sqrt{\mathrm{9}+\mathrm{x}^{\mathrm{2}} }}\:+\:\mathrm{c} \\ $$
Answered by puissant last updated on 07/Jul/21
8) ∫(((e^2 cos(x))(e^(sin(x)) ))/(e^(lne) +ln1))dx =∫((cos(x)e^(2+sin(x)) )/e)dx =(1/e)∫cos(x)e^(2+sin(x)) dx =(1/e)e^(2+sin(x)) +c...
$$\left.\mathrm{8}\right)\:\int\frac{\left(\mathrm{e}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{x}\right)\right)\left(\mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} \right)}{\mathrm{e}^{\mathrm{lne}} +\mathrm{ln1}}\mathrm{dx} \\ $$$$=\int\frac{\mathrm{cos}\left(\mathrm{x}\right)\mathrm{e}^{\mathrm{2}+\mathrm{sin}\left(\mathrm{x}\right)} }{\mathrm{e}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}\int\mathrm{cos}\left(\mathrm{x}\right)\mathrm{e}^{\mathrm{2}+\mathrm{sin}\left(\mathrm{x}\right)} \mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}\mathrm{e}^{\mathrm{2}+\mathrm{sin}\left(\mathrm{x}\right)} +\mathrm{c}… \\ $$

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