calculate ∫_0 ^1 (x^3 −2)(√(x^2 +3))dx


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Question Number 60050 by maxmathsup by imad last updated on 17/May/19

$c a l c u l a t e \int_{0}^{1} (x^{3} - 2) \sqrt{x^{2} + 3} d x$
Commented by maxmathsup by imad last updated on 18/May/19
$let A =∫_0 ^1 (x^3 −2)(√(x^2 +3))dx ⇒A =∫_0 ^1 x^3 (√(x^2 +3))dx −2 ∫_0 ^1 (√(x^2 +3))dx=H−K changement (√(x^2 +3))=t give x^2 +3 =t^2 ⇒2xdx =2tdt ⇒ A =∫_(√3) ^2 (t^2 −3)t tdt =∫_(√3) ^2 t^2 (t^2 −3)dt =∫_(√3) ^2 (t^4 −3t^2 )dt =[(1/5)t^5 −t^3 ]_(√3) ^2 =((32)/5) −8−(1/5)((√3))^5 +((√3))^3 ∫_0 ^1 (√(x^2 +3))dx =_(x =(√3)sh(t)) ∫_0 ^(argsh((1/(√3)))) (√3)ch(t)(√3)ch(t)dt =3 ∫_0 ^(ln((1/(√3)) +(√(1+(1/3))))) ch^2 (t)dt =(3/2) ∫_0 ^(ln((1/(√3))+(2/(√3)))) (1+ch(2t))dt =(3/2)ln((√3)) +(3/4)[sh(2t)]_0 ^(ln((√3))) =(3/2)ln((√3))+(3/8)[e^(2t) −e^(−2t) ]_0 ^(ln((√3))) =(3/4)ln(3) +(3/8){ ((√3))^2 −(1/(((√3))^2 ))} =(3/4)ln(3)+(3/8){3−(1/3)} =(3/4)ln(3) +1 so the value of A is determined .$
$l e t A = \int_{0}^{1} (x^{3} - 2) \sqrt{x^{2} + 3} d x \Rightarrow A = \int_{0}^{1} x^{3} \sqrt{x^{2} + 3} d x - 2 \int_{0}^{1} \sqrt{x^{2} + 3} d x = H - K$ $c h a n g e m e n t \sqrt{x^{2} + 3} = t g i v e x^{2} + 3 = t^{2} \Rightarrow 2 x d x = 2 t d t \Rightarrow$ $A = \int_{\sqrt{3}}^{2} (t^{2} - 3) t t d t = \int_{\sqrt{3}}^{2} t^{2} (t^{2} - 3) d t = \int_{\sqrt{3}}^{2} (t^{4} - 3 t^{2}) d t$ $= {[\frac{1}{5} t^{5} - t^{3}]}_{\sqrt{3}}^{2} = \frac{32}{5} - 8 - \frac{1}{5} {(\sqrt{3})}^{5} + {(\sqrt{3})}^{3}$ $\int_{0}^{1} \sqrt{x^{2} + 3} d x =_{x = \sqrt{3} s h (t)} \int_{0}^{a r g s h (\frac{1}{\sqrt{3}})} \sqrt{3} c h (t) \sqrt{3} c h (t) d t$ $= 3 \int_{0}^{l n (\frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}})} {c h}^{2} (t) d t = \frac{3}{2} \int_{0}^{l n (\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})} (1 + c h (2 t)) d t$ $= \frac{3}{2} l n (\sqrt{3}) + \frac{3}{4} {[s h (2 t)]}_{0}^{l n (\sqrt{3})} = \frac{3}{2} l n (\sqrt{3}) + \frac{3}{8} {[e^{2 t} - e^{- 2 t}]}_{0}^{l n (\sqrt{3})}$ $= \frac{3}{4} l n (3) + \frac{3}{8} {{(\sqrt{3})}^{2} - \frac{1}{{(\sqrt{3})}^{2}}} = \frac{3}{4} l n (3) + \frac{3}{8} {3 - \frac{1}{3}}$ $= \frac{3}{4} l n (3) + 1 s o t h e v a l u e o f A i s d e t e r m i n e d .$
	Answered by tanmay last updated on 17/May/19
	$∫x^3 (√(x^2 +3)) dx−2∫(√(x^2 +3)) dx I_1 =∫x^3 (√(x^2 +3)) dx t^2 =x^2 +3 2tdt=2xdx ∫x^2 ×x(√(x^2 +3)) dx ∫(t^2 −3)×t×tdt ∫(t^4 −3t^2 )dt =(t^5 /5)−t^3 +c =(((x^2 +3)^(5/2) )/5)−(x^2 +3)^(3/2) +c I_2 =∫(√(x^2 +3)) dx→use formula =((x(√(x^2 +3)))/2)+(3/2)ln(x+(√(x^2 +3)) )+c_1 I=I_1 −2I_2 I=∣(((x^2 +3)^(5/2) )/5)−(x^2 +3)^(3/2) −x(√(x^2 +3)) −3ln(x+(√(x^2 +3)) )∣_0 ^1 =(1/5){(4)^(5/2) −(3)^(5/2) }−{(4)^(3/2) −3^(3/2) }−{(√4) −0}−3ln((3/(√3))) =((32)/5)−(9/5)(√3) −8+3(√3) −2−(3/2)ln3 =((32)/5)−10+3(√3) (1−(3/5))−(3/2)ln3 =((−18)/5)+(6/5)(√3) −(3/2)ln3$
	$\int x^{3} \sqrt{x^{2} + 3} d x - 2 \int \sqrt{x^{2} + 3} d x$ $I_{1} = \int x^{3} \sqrt{x^{2} + 3} d x$ $t^{2} = x^{2} + 3$ $2 t d t = 2 x d x$ $\int x^{2} \times x \sqrt{x^{2} + 3} d x$ $\int (t^{2} - 3) \times t \times t d t$ $\int (t^{4} - 3 t^{2}) d t$ $= \frac{t^{5}}{5} - t^{3} + c$ $= \frac{{(x^{2} + 3)}^{\frac{5}{2}}}{5} - {(x^{2} + 3)}^{\frac{3}{2}} + c$ $I_{2} = \int \sqrt{x^{2} + 3} d x \to u s e f o r m u l a$ $= \frac{x \sqrt{x^{2} + 3}}{2} + \frac{3}{2} l n (x + \sqrt{x^{2} + 3}) + c_{1}$ $I = I_{1} - 2 I_{2}$ $I =∣ \frac{{(x^{2} + 3)}^{\frac{5}{2}}}{5} - {(x^{2} + 3)}^{\frac{3}{2}} - x \sqrt{x^{2} + 3} - 3 l n (x + \sqrt{x^{2} + 3}) ∣_{0}^{1}$ $= \frac{1}{5} {{(4)}^{\frac{5}{2}} - {(3)}^{\frac{5}{2}}} - {{(4)}^{\frac{3}{2}} - 3^{\frac{3}{2}}} - {\sqrt{4} - 0} - 3 l n (\frac{3}{\sqrt{3}})$ $= \frac{32}{5} - \frac{9}{5} \sqrt{3} - 8 + 3 \sqrt{3} - 2 - \frac{3}{2} l n 3$ $= \frac{32}{5} - 10 + 3 \sqrt{3} (1 - \frac{3}{5}) - \frac{3}{2} l n 3$ $= \frac{- 18}{5} + \frac{6}{5} \sqrt{3} - \frac{3}{2} l n 3$
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