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Question Number 163497 by Zaynal last updated on 07/Jan/22

Commented by Clide17 last updated on 07/Jan/22

$The function is d i s c o n t i n u o u s at interval [0, 3] .$
	Answered by TheSupreme last updated on 07/Jan/22
	$x^3 +2x^2 +x−2=(x−2)(x^2 +1) (x/((x−2)(x^2 +1)))=(A/(x−2))+((Bx+C)/(x^2 +1))=((Ax^2 +A+Bx^2 +Cx−2Bx−2C)/((x^2 +1)(x−2)))=(((A+B)x^2 +(C−2B)x+A−2C)/(...))=(x/(...)) { ((A+B=0)),((C−2B=1)),((A−2C=0)) :} → { ((A=(2/5))),((B=−(2/5))),((C=(1/5))) :} (1/5)[∫_0 ^3 (2/(x−2))dx−∫_0 ^3 ((2x)/((x^2 +1)))dx+∫_0 ^3 (1/((x^2 +1)))dx]= =(1/5)[2ln∣x−2∣−ln(x^2 +1)+arctan(x)]∣_0 ^3 =−(2/5)ln(2)−(1/5)ln(10)+(1/5)arctan(3)$
	$x^{3} + 2 x^{2} + x - 2 = (x - 2) (x^{2} + 1)$ $\frac{x}{(x - 2) (x^{2} + 1)} = \frac{A}{x - 2} + \frac{B x + C}{x^{2} + 1} = \frac{{A x}^{2} + A + {B x}^{2} + C x - 2 B x - 2 C}{(x^{2} + 1) (x - 2)} = \frac{(A + B) x^{2} + (C - 2 B) x + A - 2 C}{. . .} = \frac{x}{. . .}$ ${\begin{cases} A + B = 0 \\ C - 2 B = 1 \\ A - 2 C = 0 \end{cases} \to {\begin{cases} A = \frac{2}{5} \\ B = - \frac{2}{5} \\ C = \frac{1}{5} \end{cases}$ $\frac{1}{5} [\int_{0}^{3} \frac{2}{x - 2} d x - \int_{0}^{3} \frac{2 x}{(x^{2} + 1)} d x + \int_{0}^{3} \frac{1}{(x^{2} + 1)} d x] =$ $= \frac{1}{5} [2 l n ∣ x - 2 ∣ - l n (x^{2} + 1) + a r c t a n (x)] ∣_{0}^{3}$ $= - \frac{2}{5} l n (2) - \frac{1}{5} l n (10) + \frac{1}{5} a r c t a n (3)$
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