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Question Number 49299 by peter frank last updated on 05/Dec/18

Commented by maxmathsup by imad last updated on 05/Dec/18

$a) l e t I = \int \frac{x e^{2 x}}{{(1 + 2 x)}^{2}} d x c h a n g e m e n t 2 x = t g i v e$ $I = \int \frac{{t e}^{t}}{2 {(1 + t)}^{2}} \frac{d t}{2} = \frac{1}{4} \int \frac{t e^{t}}{{(1 + t)}^{2}} d t b u t \frac{d}{d t} (\frac{e^{t}}{1 + t}) = \frac{e^{t} (1 + t) - e^{t}}{{(1 + t)}^{2}} = \frac{t e^{t}}{{(1 + t)}^{2}} \Rightarrow$ $I = \frac{1}{4} \frac{e^{t}}{t + 1} + c = \frac{e^{2 x}}{4 (2 x + 1)} + c .$
Commented by Abdo msup. last updated on 05/Dec/18
$b) let A =∫_1 ^2 (dx/(x^2 (√(5x^2 −1)))) ⇒A =(1/(√5))∫_1 ^2 (dx/(x^2 (√(x^2 −(1/5))))) =_(x =(1/(√5))ch(t)) (1/(5(√5))) ∫_(argch((√5))) ^(argch(2(√5))) ((sh(t)dt)/((1/5)ch^2 tsh(t))) =(1/(√5)) ∫_(ln((√5)+2)) ^(ln(2(√5)+(√(19)))) ((2dt)/(1+ch(2t))) =(2/(√5)) ∫_(ln(2+(√5))) ^(ln((√(19)) +2(√5){) (dt/(1+((e^(2t) +e^(−2t) )/2))) =(4/(√5)) ∫_(ln(2+(√5))) ^(ln((√(19)) +2(√5))) (dt/(2 +e^(2t) +e^(−2t) )) =_(e^(2t) =u) (4/(√5)) ∫_((2+(√5))^2 ) ^(((√(19))+2(√5))^2 ) (du/(2(2 +u +u^(−1) )u)) =(2/(√5)) ∫_((2+(√5))^2 ) ^(((√(19)) +2(√5))^2 ) (du/(2u +u^2 +1)) =(2/(√5)) ∫_((2+(√5))^2 ) ^(((√(19))+2(√5))^2 ) (du/((u+1)^2 )) =(2/(√5))[−(1/(1+u))]_((2+(√5))^2 ) ^(((√(19))+2(√5))^2 ) =(2/(√5)){ (1/(1+(2+(√5))^2 )) −(1/(1+((√(19))+2(√5))^2 ))} . =$
$b) l e t A = \int_{1}^{2} \frac{d x}{x^{2} \sqrt{5 x^{2} - 1}} \Rightarrow A = \frac{1}{\sqrt{5}} \int_{1}^{2} \frac{d x}{x^{2} \sqrt{x^{2} - \frac{1}{5}}}$ $=_{x = \frac{1}{\sqrt{5}} c h (t)} \frac{1}{5 \sqrt{5}} \int_{a r g c h (\sqrt{5})}^{a r g c h (2 \sqrt{5})} \frac{s h (t) d t}{\frac{1}{5} {c h}^{2} t s h (t)}$ $= \frac{1}{\sqrt{5}} \int_{l n (\sqrt{5} + 2)}^{l n (2 \sqrt{5} + \sqrt{19})} \frac{2 d t}{1 + c h (2 t)}$ $= \frac{2}{\sqrt{5}} \int_{l n (2 + \sqrt{5})}^{l n (\sqrt{19} + 2 \sqrt{5} {} \frac{d t}{1 + \frac{e^{2 t} + e^{- 2 t}}{2}}$ $= \frac{4}{\sqrt{5}} \int_{l n (2 + \sqrt{5})}^{l n (\sqrt{19} + 2 \sqrt{5})} \frac{d t}{2 + e^{2 t} + e^{- 2 t}}$ $=_{e^{2 t} = u} \frac{4}{\sqrt{5}} \int_{{(2 + \sqrt{5})}^{2}}^{{(\sqrt{19} + 2 \sqrt{5})}^{2}} \frac{d u}{2 (2 + u + u^{- 1}) u}$ $= \frac{2}{\sqrt{5}} \int_{{(2 + \sqrt{5})}^{2}}^{{(\sqrt{19} + 2 \sqrt{5})}^{2}} \frac{d u}{2 u + u^{2} + 1}$ $= \frac{2}{\sqrt{5}} \int_{{(2 + \sqrt{5})}^{2}}^{{(\sqrt{19} + 2 \sqrt{5})}^{2}} \frac{d u}{{(u + 1)}^{2}} = \frac{2}{\sqrt{5}} {[- \frac{1}{1 + u}]}_{{(2 + \sqrt{5})}^{2}}^{{(\sqrt{19} + 2 \sqrt{5})}^{2}}$ $= \frac{2}{\sqrt{5}} {\frac{1}{1 + {(2 + \sqrt{5})}^{2}} - \frac{1}{1 + {(\sqrt{19} + 2 \sqrt{5})}^{2}}} .$ $=$
Commented by peter frank last updated on 06/Dec/18

$thank you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
	$V=πr^2 h (dV/dt)=πr^2 (dh/dt)+πh×2r×(dr/dt) (dV/dt)=π{8^2 ×(−0.5)+2×8×12×0.2} =π(−32+38.4) =6.4×π$
	$V = π r^{2} h$ $\frac{d V}{d t} = π r^{2} \frac{d h}{d t} + π h \times 2 r \times \frac{d r}{d t}$ $\frac{d V}{d t} = π {8^{2} \times (- 0.5) + 2 \times 8 \times 12 \times 0.2}$ $= π (- 32 + 38.4)$ $= 6.4 \times π$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

	$\int \frac{{x e}^{2 x}}{{(1 + 2 x)}^{2}} d x$ $= \frac{1}{2} \int \frac{(1 + 2 x - 1) e^{2 x}}{{(1 + 2 x)}^{2}} d x$ $= \frac{1}{2} \int \frac{(1 + 2 x) e^{2 x} - e^{2 x}}{{(1 + 2 x)}^{2}} d x$ $= \frac{1}{4} \int \frac{(1 + 2 x) \frac{d (e^{2 x})}{d x} - e^{2 x} \times \frac{d}{d x} (1 + 2 x)}{{(1 + 2 x)}^{2}} d x$ $= \frac{1}{4} \int \frac{d}{d x} (\frac{e^{2 x}}{1 + 2 x}) d x$ $= \frac{1}{4} \int d (\frac{e^{2 x}}{1 + 2 x})$ $= \frac{1}{4} \times \frac{e^{2 x}}{1 + 2 x} + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

	$\int_{1}^{2} \frac{d x}{x^{2} \sqrt{5 x^{2} - 1}}$ $I = \int \frac{d x}{x^{2} \sqrt{5 x^{2} - 1}} x = \frac{1}{t} d x = \frac{- d t}{t^{2}}$ $\int \frac{- d t}{t^{2} \times \frac{1}{t^{2}} \times \sqrt{5 {(\frac{1}{t})}^{2} - 1}}$ $\int \frac{- t \times d t}{\sqrt{5 - t^{2}}}$ $\frac{1}{2} \int \frac{d (5 - t^{2})}{\sqrt{5 - t^{2}}}$ $= \frac{1}{2} \times \frac{{(5 - t^{2})}^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} + c$ $= \sqrt{5 - t^{2}} + c$ $= \sqrt{5 - \frac{1}{x^{2}}} + c$ $s o a n s w e r o f r e s u i r e d i n t r e g a l i s$ $∣ \sqrt{5 - \frac{1}{x^{2}}} ∣_{1}^{2}$ $= \sqrt{5 - \frac{1}{4}} - \sqrt{5 - \frac{1}{1}}$ $= \frac{\sqrt{19}}{2} - 2$
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