Find ∫_(−1) ^1 ((9 +4x^2 )/(9−4x^2 )) dx


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Question Number 66105 by Rio Michael last updated on 09/Aug/19

$F i n d \int_{- 1}^{1} \frac{9 + 4 x^{2}}{9 - 4 x^{2}} d x$
Commented by Prithwish sen last updated on 09/Aug/19

$\int_{- 1}^{1} \frac{18 - (9 - {4 x}^{2})}{9 - {4 x}^{2}} dx = 2 \int_{0}^{1} [\frac{18}{9 - {4 x}^{2}} - 1] dx$ $= 2 {[\int \frac{18 dx}{9 - {4 x}^{2}} - \int dx]}_{0}^{1} for the first part use \int \frac{dx}{a^{2} - x^{2}} = \frac{1}{2 a} \int \frac{1}{a - x} + \frac{1}{a + x} dx$ $= 3 \ln 5 - 2$ $please check the calcultion .$
Commented by mathmax by abdo last updated on 09/Aug/19
$let I =∫_(−1) ^1 ((9+4x^2 )/(9−4x^2 ))dx ⇒ I =2 ∫_0 ^1 ((4x^2 +9)/(−4x^2 +9))dx =−2 ∫_0 ^1 ((4x^2 +9)/(4x^2 −9))dx =−2 ∫_0 ^1 ((4x^2 −9 +18)/(4x^2 −9))dx =−2 −36 ∫_0 ^1 (dx/((2x−3)(2x+3))) =−2 −6 ∫_0 ^1 {(1/(2x−3))−(1/(2x+3))}dx =−2−6[ln∣((2x−3)/(2x+3))∣]_0 ^1 =−2 −6{ln((1/5))−0}=−2+6ln(5) ⇒ I =6ln(5)−2$
$l e t I = \int_{- 1}^{1} \frac{9 + 4 x^{2}}{9 - 4 x^{2}} d x \Rightarrow I = 2 \int_{0}^{1} \frac{4 x^{2} + 9}{- 4 x^{2} + 9} d x = - 2 \int_{0}^{1} \frac{4 x^{2} + 9}{4 x^{2} - 9} d x$ $= - 2 \int_{0}^{1} \frac{4 x^{2} - 9 + 18}{4 x^{2} - 9} d x = - 2 - 36 \int_{0}^{1} \frac{d x}{(2 x - 3) (2 x + 3)}$ $= - 2 - 6 \int_{0}^{1} {\frac{1}{2 x - 3} - \frac{1}{2 x + 3}} d x$ $= - 2 - 6 {[l n ∣ \frac{2 x - 3}{2 x + 3} ∣]}_{0}^{1} = - 2 - 6 {l n (\frac{1}{5}) - 0} = - 2 + 6 l n (5) \Rightarrow$ $I = 6 l n (5) - 2$
Commented by Rio Michael last updated on 09/Aug/19

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