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Question Number 106705 by bemath last updated on 06/Aug/20

@bemath@ ∫ ((1+x^2 )/(√(9−4x^2 ))) dx

$$\:\:\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\:\:\:\:\:\int\:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\sqrt{\mathrm{9}−\mathrm{4x}^{\mathrm{2}} }}\:\mathrm{dx}\: \\ $$

Answered by mathmax by abdo last updated on 06/Aug/20

I =∫ ((1+x^2 )/(√(9−4x^2 )))dx ⇒I =∫ ((1+x^2 )/(3(√(1−(4/9)x^2 )))) we do the changement (2/3)x =sint ⇒ I =∫ ((1+(9/4)sin^2 t)/(3 cost))×(3/2) cost dt =(1/2) ∫ ((4+9sin^2 t)/4) dt =(1/2)t +(9/8) ∫ sin^2 t dt =(t/2) +(9/(16))∫(1−cos(2t))dt =(t/2) +((9t)/(16))−(9/(32))sin(2t) +C =((17t)/(16)) −(9/(16))sint cost +C =((17)/(16)) arcsin(((2x)/3))−(9/(16)).((2x)/3)(√(1−(4/9)x^2 )) +C I=((17)/(16))arcsin(((2x)/3))−((3x)/8)(√(1−(4/9)x^2 )) +C

$$\mathrm{I}\:=\int\:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\sqrt{\mathrm{9}−\mathrm{4x}^{\mathrm{2}} }}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int\:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}\mathrm{x}^{\mathrm{2}} }}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\: \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}\:=\mathrm{sint}\:\Rightarrow\:\mathrm{I}\:=\int\:\:\frac{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{3}\:\mathrm{cost}}×\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{cost}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{\mathrm{4}+\mathrm{9sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{4}}\:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}\:+\frac{\mathrm{9}}{\mathrm{8}}\:\int\:\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{t}}{\mathrm{2}}\:+\frac{\mathrm{9}}{\mathrm{16}}\int\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{2t}\right)\right)\mathrm{dt}\:=\frac{\mathrm{t}}{\mathrm{2}}\:+\frac{\mathrm{9t}}{\mathrm{16}}−\frac{\mathrm{9}}{\mathrm{32}}\mathrm{sin}\left(\mathrm{2t}\right)\:+\mathrm{C} \\ $$$$=\frac{\mathrm{17t}}{\mathrm{16}}\:−\frac{\mathrm{9}}{\mathrm{16}}\mathrm{sint}\:\mathrm{cost}\:+\mathrm{C} \\ $$$$=\frac{\mathrm{17}}{\mathrm{16}}\:\mathrm{arcsin}\left(\frac{\mathrm{2x}}{\mathrm{3}}\right)−\frac{\mathrm{9}}{\mathrm{16}}.\frac{\mathrm{2x}}{\mathrm{3}}\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}\mathrm{x}^{\mathrm{2}} }\:+\mathrm{C} \\ $$$$\mathrm{I}=\frac{\mathrm{17}}{\mathrm{16}}\mathrm{arcsin}\left(\frac{\mathrm{2x}}{\mathrm{3}}\right)−\frac{\mathrm{3x}}{\mathrm{8}}\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}\mathrm{x}^{\mathrm{2}} }\:+\mathrm{C} \\ $$

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