Question Number 66105 by Rio Michael last updated on 09/Aug/19
$${Find}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{9}\:+\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }\:{dx} \\ $$
Commented by Prithwish sen last updated on 09/Aug/19
![∫_(−1) ^1 ((18−(9−4x^2 ) )/(9−4x^2 )) dx = 2∫_0 ^1 [((18)/(9−4x^2 )) −1] dx = 2[∫((18dx)/(9−4x^2 )) −∫dx]_0 ^1 for the first part use ∫(dx/(a^2 −x^2 )) = (1/(2a))∫(1/(a−x)) +(1/(a+x)) dx = 3ln5−2 please check the calcultion.](Q66121.png)
$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{18}−\left(\mathrm{9}−\mathrm{4x}^{\mathrm{2}} \right)\:}{\mathrm{9}−\mathrm{4x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\mathrm{18}}{\mathrm{9}−\mathrm{4x}^{\mathrm{2}} }\:−\mathrm{1}\right]\:\mathrm{dx} \\ $$$$=\:\mathrm{2}\left[\int\frac{\mathrm{18dx}}{\mathrm{9}−\mathrm{4x}^{\mathrm{2}} }\:−\int\mathrm{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{for}\:\mathrm{the}\:\mathrm{first}\:\mathrm{part}\:\mathrm{use}\:\int\frac{\mathrm{dx}}{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2a}}\int\frac{\mathrm{1}}{\mathrm{a}−\mathrm{x}}\:+\frac{\mathrm{1}}{\mathrm{a}+\mathrm{x}}\:\mathrm{dx}\: \\ $$$$=\:\mathrm{3ln5}−\mathrm{2} \\ $$$$\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{calcultion}. \\ $$
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](Q66142.png)
$${let}\:{I}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{9}+\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }{dx}\:\Rightarrow\:{I}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{9}}{−\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{9}}{dx}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{9}}{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}}{dx} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}\:+\mathrm{18}}{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}}{dx}\:=−\mathrm{2}\:−\mathrm{36}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left(\mathrm{2}{x}−\mathrm{3}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)} \\ $$$$=−\mathrm{2}\:−\mathrm{6}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{3}}\right\}{dx}\: \\ $$$$=−\mathrm{2}−\mathrm{6}\left[{ln}\mid\frac{\mathrm{2}{x}−\mathrm{3}}{\mathrm{2}{x}+\mathrm{3}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\mathrm{2}\:−\mathrm{6}\left\{{ln}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{0}\right\}=−\mathrm{2}+\mathrm{6}{ln}\left(\mathrm{5}\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{6}{ln}\left(\mathrm{5}\right)−\mathrm{2} \\ $$
Commented by Rio Michael last updated on 09/Aug/19
$${thank}\:{you}\: \\ $$