Given that g(x)=(2/((1+x)(1+3x^2 )) a) express g(x) in partial fractions. b) evaluate ∫


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Question Number 64762 by Lontum Hans-Sandys last updated on 21/Jul/19
$Given that g(x)=(2/((1+x)(1+3x^2 )) a) express g(x) in partial fractions. b) evaluate ∫_0 ^1 g((x) dx.$
$Given that g (x) = \frac{2}{(1 + x) (1 + {3 x}^{2}}$ $a) express g (x) in partial fractions .$ $b) evaluate \int_{0}^{1} g ((x) dx .$
Commented by mathmax by abdo last updated on 21/Jul/19
$1)g(x) =(a/(x+1)) +((bx+c)/(3x^2 +1)) a =lim_(x→−1) (x+1)g(x) =(1/2) lim_(x→+∞) xg(x) =0 =a+(b/3) ⇒3a+b =0 ⇒b=−(3/2) ⇒ g(x) =(1/(2(x+1))) −(1/2) ((3x−2c)/(3x^2 +1)) g(0) =2 =(1/2) +c ⇒c =2−(1/2) =(3/2) ⇒g(x) =(1/(2(x+1)))−(1/2)((3x−3)/(3x^2 +1)) b) ∫_0 ^1 g(x)dx =(1/2) ∫_0 ^1 (dx/(x+1)) −(1/2) ∫_0 ^1 ((x−1)/((x^2 +(1/3))))dx ∫_0 ^1 (dx/(x+1)) =[ln∣x+1∣]_0 ^1 =ln(2) ∫_0 ^1 ((x−1)/(x^2 +(1/3)))dx =(1/2) ∫_0 ^1 ((2x)/(x^2 +(1/3)))dx −∫_0 ^1 (dx/(x^2 +(1/3))) =[ln(x^2 +(1/3))]_0 ^1 −∫_0 ^1 (dx/(x^2 +(1/3))) =ln((4/3))−ln((1/3))−∫_0 ^1 (dx/(x^2 +(1/3))) =2ln(2)−∫_0 ^1 (dx/(x^2 +(1/3))) but we have ∫_0 ^1 (dx/(x^2 +(1/3))) =_(x =(t/(√3))) =∫_0 ^(√3) (dt/(((√3)/3)(1+t^2 ))) =(√3)[ arctan(t)]_0 ^(√3) =(√3) arctan((√3))=((π(√3))/3) ⇒ ∫_0 ^1 g(x)dx =((ln(2))/2) −(1/2){2ln(2)−((π(√3))/3)} =−((ln(2))/2) +((π(√3))/6)$
$1) g (x) = \frac{a}{x + 1} + \frac{b x + c}{3 x^{2} + 1}$ $a = {l i m}_{x \to - 1} (x + 1) g (x) = \frac{1}{2}$ ${l i m}_{x \to + \infty} x g (x) = 0 = a + \frac{b}{3} \Rightarrow 3 a + b = 0 \Rightarrow b = - \frac{3}{2} \Rightarrow$ $g (x) = \frac{1}{2 (x + 1)} - \frac{1}{2} \frac{3 x - 2 c}{3 x^{2} + 1}$ $g (0) = 2 = \frac{1}{2} + c \Rightarrow c = 2 - \frac{1}{2} = \frac{3}{2} \Rightarrow g (x) = \frac{1}{2 (x + 1)} - \frac{1}{2} \frac{3 x - 3}{3 x^{2} + 1}$ $b) \int_{0}^{1} g (x) d x = \frac{1}{2} \int_{0}^{1} \frac{d x}{x + 1} - \frac{1}{2} \int_{0}^{1} \frac{x - 1}{(x^{2} + \frac{1}{3})} d x$ $\int_{0}^{1} \frac{d x}{x + 1} = {[l n ∣ x + 1 ∣]}_{0}^{1} = l n (2)$ $\int_{0}^{1} \frac{x - 1}{x^{2} + \frac{1}{3}} d x = \frac{1}{2} \int_{0}^{1} \frac{2 x}{x^{2} + \frac{1}{3}} d x - \int_{0}^{1} \frac{d x}{x^{2} + \frac{1}{3}}$ $= {[l n (x^{2} + \frac{1}{3})]}_{0}^{1} - \int_{0}^{1} \frac{d x}{x^{2} + \frac{1}{3}} = l n (\frac{4}{3}) - l n (\frac{1}{3}) - \int_{0}^{1} \frac{d x}{x^{2} + \frac{1}{3}}$ $= 2 l n (2) - \int_{0}^{1} \frac{d x}{x^{2} + \frac{1}{3}} b u t w e h a v e \int_{0}^{1} \frac{d x}{x^{2} + \frac{1}{3}} =_{x = \frac{t}{\sqrt{3}}}$ $= \int_{0}^{\sqrt{3}} \frac{d t}{\frac{\sqrt{3}}{3} (1 + t^{2})} = \sqrt{3} {[a r c t a n (t)]}_{0}^{\sqrt{3}} = \sqrt{3} a r c t a n (\sqrt{3}) = \frac{π \sqrt{3}}{3} \Rightarrow$ $\int_{0}^{1} g (x) d x = \frac{l n (2)}{2} - \frac{1}{2} {2 l n (2) - \frac{π \sqrt{3}}{3}} = - \frac{l n (2)}{2} + \frac{π \sqrt{3}}{6}$
	Answered by Tanmay chaudhury last updated on 21/Jul/19

	$\frac{2}{(1 + x) (1 + 3 x^{2})} = \frac{a}{1 + x} + \frac{b x + c}{1 + 3 x^{2}}$ $2 = a (1 + 3 x^{2}) + (1 + x) (b x + c)$ $2 = a + 3 {a x}^{2} + b x + c + {b x}^{2} + c x$ $2 = x^{2} (3 a + b) + x (b + c) + (a + c)$ $3 a + b = 0$ $b + c = 0$ $s o 3 a - c = 0$ $c = 3 a$ $a + c = 2$ $3 a + a = 2 \to a = \frac{2}{4} = \frac{1}{2} a n d c = \frac{3}{2}$ $b = - c$ $b = \frac{- 3}{2}$ $h e n c e \frac{2}{(1 + x) ((1 + 3 x^{2})} = \frac{\frac{1}{2}}{x + 1} + \frac{\frac{- 3 x}{2} + \frac{3}{2}}{1 + 3 x^{2}}$ $\int \frac{2}{(1 + x) (1 + 3 x^{2})} = \int \frac{\frac{1}{2}}{x + 1} d x + (\frac{- 3}{2}) \int \frac{x d x}{1 + 3 x^{2}} + \frac{3}{2} \int \frac{d x}{1 + 3 x^{2}}$ $= \frac{1}{2} \int \frac{d x}{x + 1} + \frac{- 3}{2 \times 6} \int \frac{6 x}{1 + 3 x^{2}} d x + \frac{1}{2} \int \frac{d x}{x^{2} + \frac{1}{3}}$ $= \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{4} \int \frac{d (1 + 3 x^{2})}{1 + 3 x^{2}} + \frac{1}{2} \int \frac{d x}{x^{2} + {(\sqrt{\frac{1}{3}})}^{2}}$ $= \frac{1}{2} l n (x + 1) - \frac{1}{4} l n (1 + 3 x^{2}) + \frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{3}}} {t a n}^{- 1} (\frac{x}{\frac{1}{\sqrt{3}}})$ $= \frac{1}{2} l n (x + 1) - \frac{1}{4} l n (1 + 3 x^{2}) + \frac{\sqrt{3}}{2} {t a n}^{- 1} (x \sqrt{3})$ $s o \int_{0}^{1} g (x) d x$ $=∣ \frac{1}{2} l n (x + 1) - \frac{1}{4} l n (1 + 3 x^{2}) + \frac{\sqrt{3}}{2} {t a n}^{- 1} (x \sqrt{3}) ∣_{0 =}^{1}$ $= \frac{1}{2} l n 2 - \frac{1}{4} l n (4) + \frac{\sqrt{3}}{2} \times \frac{π}{3}$ $= \frac{π}{2 \sqrt{3}}$
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