find ∫_0 ^1 ((1+x^2 )/(1+x^3 ))dx


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Question Number 42792 by maxmathsup by imad last updated on 02/Sep/18

$f i n d \int_{0}^{1} \frac{1 + x^{2}}{1 + x^{3}} d x$
Commented by behi83417@gmail.com last updated on 03/Sep/18

$I = \frac{1}{3} l n (1 + x^{3}) + A . l n (1 + x) + B . l n \frac{x - a}{x - b} + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18
	$∫_0 ^1 (dx/(1+x^3 ))+(1/3)∫_0 ^1 ((3x^2 )/(1+x^3 )) ∫_0 ^1 (dx/((1+x)(1−x+x^2 )))+(1/3)∫_0 ^1 ((d(1+x^3 ))/(1+x^3 )) I_2 =(1/3)∫_0 ^1 ((d(1+x^3 ))/(1+x^3 ))=(1/3)∣ln(1+x^3 )∣_0 ^1 =(1/3)ln2 now (1/((1+x)(1−x+x^2 )))=(a/(1+x))+((bx+c)/(x^2 −x+1)) 1=a(x^2 −x+1)+(bx+c)(1+x) 1=ax^2 −ax+a+bx+bx^2 +c+cx contd 1=x^2 (a+b)+x(−a+b+c)+(a+c) a+b=0 −a+b+c=0 a+c=1 −a−a+c=0 c=2a a+2a=1 a=(1/3) b=((−1)/3) c=(2/3) I_1 =∫_0 ^1 {(a/(1+x))+((bx+c)/(x^2 −x+1)) }dx =(1/3)∫_(0 ) ^1 (dx/(1+x))+(1/3)∫_0 ^1 ((−x+2)/(x^2 −x+1))dx =(1/3)∫_0 ^1 (dx/(1+x))−(1/6)∫_0 ^1 ((2x−1−3)/(x^2 −x+1))dx =(1/3)ln2−(1/6)∫_0 ^1 ((d(x^2 −x+1))/(x^2 −x+1))+(1/2)∫_0 ^1 (dx/((x^2 −2.x.(1/2)+(1/4)+1−(1/4)))) =(1/3)ln2−(1/6)∣ln(x^2 −x+1)∣_0 ^1 +(1/2)∫_0 ^1 (dx/((x−(1/2))^2 +(((√3)/2))^2 )) =(1/3)ln2−(1/6)×0+(1/2)×(2/(√3))∣tan^(−1) (((x−(1/2))/((√3)/2)))∣_0 ^1 =(1/3)ln2+(1/(√3)){tan^(−1) (1/(√3))−tan^(−1) (−(1/(√3)))} =(1/3)ln2+(2/(√3))×(Π/6)=(1/3)ln2+(Π/(3(√3))) clmplete ans is =(2/3)ln2+(Π/(3(√3)))$
	$\int_{0}^{1} \frac{d x}{1 + x^{3}} + \frac{1}{3} \int_{0}^{1} \frac{3 x^{2}}{1 + x^{3}}$ $\int_{0}^{1} \frac{d x}{(1 + x) (1 - x + x^{2})} + \frac{1}{3} \int_{0}^{1} \frac{d (1 + x^{3})}{1 + x^{3}}$ $I_{2} = \frac{1}{3} \int_{0}^{1} \frac{d (1 + x^{3})}{1 + x^{3}} = \frac{1}{3} ∣ l n (1 + x^{3}) ∣_{0}^{1} = \frac{1}{3} l n 2$ $n o w$ $\frac{1}{(1 + x) (1 - x + x^{2})} = \frac{a}{1 + x} + \frac{b x + c}{x^{2} - x + 1}$ $1 = a (x^{2} - x + 1) + (b x + c) (1 + x)$ $1 = {a x}^{2} - a x + a + b x + {b x}^{2} + c + c x$ $c o n t d$ $1 = x^{2} (a + b) + x (- a + b + c) + (a + c)$ $a + b = 0 - a + b + c = 0 a + c = 1$ $- a - a + c = 0 c = 2 a a + 2 a = 1$ $a = \frac{1}{3} b = \frac{- 1}{3} c = \frac{2}{3}$ $I_{1} = \int_{0}^{1} {\frac{a}{1 + x} + \frac{b x + c}{x^{2} - x + 1}} d x$ $= \frac{1}{3} \int_{0}^{1} \frac{d x}{1 + x} + \frac{1}{3} \int_{0}^{1} \frac{- x + 2}{x^{2} - x + 1} d x$ $= \frac{1}{3} \int_{0}^{1} \frac{d x}{1 + x} - \frac{1}{6} \int_{0}^{1} \frac{2 x - 1 - 3}{x^{2} - x + 1} d x$ $= \frac{1}{3} l n 2 - \frac{1}{6} \int_{0}^{1} \frac{d (x^{2} - x + 1)}{x^{2} - x + 1} + \frac{1}{2} \int_{0}^{1} \frac{d x}{(x^{2} - 2 . x . \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4})}$ $= \frac{1}{3} l n 2 - \frac{1}{6} ∣ l n (x^{2} - x + 1) ∣_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$ $= \frac{1}{3} l n 2 - \frac{1}{6} \times 0 + \frac{1}{2} \times \frac{2}{\sqrt{3}} ∣ {t a n}^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) ∣_{0}^{1}$ $= \frac{1}{3} l n 2 + \frac{1}{\sqrt{3}} {{t a n}^{- 1} \frac{1}{\sqrt{3}} - {t a n}^{- 1} (- \frac{1}{\sqrt{3}})}$ $= \frac{1}{3} l n 2 + \frac{2}{\sqrt{3}} \times \frac{Π}{6} = \frac{1}{3} l n 2 + \frac{Π}{3 \sqrt{3}}$ $c l m p l e t e a n s i s = \frac{2}{3} l n 2 + \frac{Π}{3 \sqrt{3}}$
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