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1-lt-a-b-then-find-a-b-tan-1-3x-1-2x-2-dx-




Question Number 145231 by mathdanisur last updated on 03/Jul/21
1<a≤b  then find  ∫_( a) ^( b)  tan^(-1) (((3x)/(1-2x^2 )))dx=?
$$\mathrm{1}<{a}\leqslant{b}\:\:{then}\:{find} \\ $$$$\underset{\:\boldsymbol{{a}}} {\overset{\:\boldsymbol{{b}}} {\int}}\:{tan}^{-\mathrm{1}} \left(\frac{\mathrm{3}{x}}{\mathrm{1}-\mathrm{2}{x}^{\mathrm{2}} }\right){dx}=? \\ $$
Commented by Dwaipayan Shikari last updated on 03/Jul/21
tan^(−1) ((3x)/(1−2x^2 ))=tan^(−1) x+tan^(−1) 2x  ∫tan^(−1) x dx=xtan^(−1) x−∫(x/(1+x^2 ))dx=xtan^(−1) x−(1/2)log(1+x^2 )  Now put these...
$${tan}^{−\mathrm{1}} \frac{\mathrm{3}{x}}{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} }={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} \mathrm{2}{x} \\ $$$$\int{tan}^{−\mathrm{1}} {x}\:{dx}={xtan}^{−\mathrm{1}} {x}−\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}={xtan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${Now}\:{put}\:{these}… \\ $$
Commented by mathdanisur last updated on 03/Jul/21
thank you Ser, answer?
$${thank}\:{you}\:{Ser},\:{answer}? \\ $$
Commented by Dwaipayan Shikari last updated on 03/Jul/21
b(tan^(−1) b+tan^(−1) 2b)−a(tan^(−1) a+tan^(−1) 2a)−(1/2)log(((1+b^2 )/(1+a^2 )).(√((1+4a^2  )/(1+4b^2 ))))
$${b}\left({tan}^{−\mathrm{1}} {b}+{tan}^{−\mathrm{1}} \mathrm{2}{b}\right)−{a}\left({tan}^{−\mathrm{1}} {a}+{tan}^{−\mathrm{1}} \mathrm{2}{a}\right)−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}+{b}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} }.\sqrt{\frac{\mathrm{1}+\mathrm{4}{a}^{\mathrm{2}} \:}{\mathrm{1}+\mathrm{4}{b}^{\mathrm{2}} }}\right) \\ $$
Answered by mathmax by abdo last updated on 04/Jul/21
Ψ=∫_a ^b  arctan(((3x)/(1−2x^2 )))dx  by parts  Ψ=[xarctan(((3x)/(1−2x^2 )))]_a ^b  −∫_a ^b  x×(((((3x)/(1−2x^2 )))^′ )/(1+(((3x)/(1−2x^2 )))^2 ))dx  but (....)^′  =((3(1−2x^2 )−3x(−4x))/((1−2x^2 )^2 ))=((3−6x^2 +12x^2 )/((1−2x^2 )^2 ))=((3+6x^2 )/((1−2x^2 )^2 )) ⇒  Ψ=barctan(((3b)/(1−2b^2 )))−aarctan(((3a)/(1−2a^2 )))  −∫_a ^b  ((3x+6x^3 )/((1−2x^2 )^2 (1+((9x^2 )/((1−2x^2 )^2 )))))dx  =.....−∫_a ^b  ((3x+6x^3 )/((1−2x^2 )^2  +9x^2 ))dx we have  ∫_a ^b  ((6x^3  +3x)/(4x^4 −4x^2  +1+9x^2 ))dx =∫_a ^b  ((6x^3  +3x)/(4x^4 +5x^2  +1))dx  4x^4  +5x^2  +1=0⇒4u^2  +5u +1=0  (u=x^2 )  →Δ=25−16=9 ⇒u_1 =((−5+3)/8)=−(1/4)and u_2 =((−5−3)/8)=−1 ⇒  4u^2  +5u+1=4(u+(1/4))(u+1) =(u+1)(4u+1)=(x^2 +1)(4x^2  +1)  let decompose F(x)=((6x^3  +3x)/((x^2  +1)(4x^2  +1))) ⇒  F(x) =((αx+β)/(x^2  +1))+((cx+d)/(4x^2  +1))  determinatin of coefficient is eazy ⇒  ∫ F(x)dx =(α/2)log(x^2  +1)+βarctanx +(c/8)log(4x^2  +1)+d∫ (dx/(4x^2  +1))  ∫ (dx/(4x^2  +1))=_(2x=t)   (1/2) ∫ (dt/(t^2  +1))=(1/2)arctan(2x)+c....
$$\Psi=\int_{\mathrm{a}} ^{\mathrm{b}} \:\mathrm{arctan}\left(\frac{\mathrm{3x}}{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} }\right)\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\Psi=\left[\mathrm{xarctan}\left(\frac{\mathrm{3x}}{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} }\right)\right]_{\mathrm{a}} ^{\mathrm{b}} \:−\int_{\mathrm{a}} ^{\mathrm{b}} \:\mathrm{x}×\frac{\left(\frac{\mathrm{3x}}{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} }\right)^{'} }{\mathrm{1}+\left(\frac{\mathrm{3x}}{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\mathrm{but}\:\left(….\right)^{'} \:=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)−\mathrm{3x}\left(−\mathrm{4x}\right)}{\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{3}−\mathrm{6x}^{\mathrm{2}} +\mathrm{12x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{3}+\mathrm{6x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\Psi=\mathrm{barctan}\left(\frac{\mathrm{3b}}{\mathrm{1}−\mathrm{2b}^{\mathrm{2}} }\right)−\mathrm{aarctan}\left(\frac{\mathrm{3a}}{\mathrm{1}−\mathrm{2a}^{\mathrm{2}} }\right) \\ $$$$−\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{3x}+\mathrm{6x}^{\mathrm{3}} }{\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{9x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)}\mathrm{dx} \\ $$$$=…..−\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{3x}+\mathrm{6x}^{\mathrm{3}} }{\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{9x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{6x}^{\mathrm{3}} \:+\mathrm{3x}}{\mathrm{4x}^{\mathrm{4}} −\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}+\mathrm{9x}^{\mathrm{2}} }\mathrm{dx}\:=\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{6x}^{\mathrm{3}} \:+\mathrm{3x}}{\mathrm{4x}^{\mathrm{4}} +\mathrm{5x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$\mathrm{4x}^{\mathrm{4}} \:+\mathrm{5x}^{\mathrm{2}} \:+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{4u}^{\mathrm{2}} \:+\mathrm{5u}\:+\mathrm{1}=\mathrm{0}\:\:\left(\mathrm{u}=\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\rightarrow\Delta=\mathrm{25}−\mathrm{16}=\mathrm{9}\:\Rightarrow\mathrm{u}_{\mathrm{1}} =\frac{−\mathrm{5}+\mathrm{3}}{\mathrm{8}}=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{and}\:\mathrm{u}_{\mathrm{2}} =\frac{−\mathrm{5}−\mathrm{3}}{\mathrm{8}}=−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{4u}^{\mathrm{2}} \:+\mathrm{5u}+\mathrm{1}=\mathrm{4}\left(\mathrm{u}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{u}+\mathrm{1}\right)\:=\left(\mathrm{u}+\mathrm{1}\right)\left(\mathrm{4u}+\mathrm{1}\right)=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{6x}^{\mathrm{3}} \:+\mathrm{3x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\alpha\mathrm{x}+\beta}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}+\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}}\:\:\mathrm{determinatin}\:\mathrm{of}\:\mathrm{coefficient}\:\mathrm{is}\:\mathrm{eazy}\:\Rightarrow \\ $$$$\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\:=\frac{\alpha}{\mathrm{2}}\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)+\beta\mathrm{arctanx}\:+\frac{\mathrm{c}}{\mathrm{8}}\mathrm{log}\left(\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}\right)+\mathrm{d}\int\:\frac{\mathrm{dx}}{\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\int\:\frac{\mathrm{dx}}{\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}}=_{\mathrm{2x}=\mathrm{t}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{2x}\right)+\mathrm{c}…. \\ $$$$ \\ $$

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