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9x-2-49-x-3-dx-




Question Number 212456 by Hanuda354 last updated on 14/Oct/24
∫  ((√(9x^2 −49))/x^3 )  dx = ?
9x249x3dx=?
Answered by Sutrisno last updated on 14/Oct/24
∫((√((3x)^2 −7^2 ))/x^3 )dx  3x=7secθ→dx=(7/3)secθtanθdθ  =∫((√((7secθ)^2 −7^2 ))/(((7/3)secθ)^3 )).(7/3)secθtanθdθ  =∫((7tanθ)/(((343)/(27))sec^3 θ)).(7/3)secθtanθdθ  =(9/7)∫((tan^2 θ)/(sec^2 θ))dθ  =(9/7)∫sin^2 θdθ  =(9/7)∫((1−cos2θ)/2)dθ  =(9/(14))(θ−(1/2)sin2θ)  =(9/(14))(θ−sinθcosθ)+c  =(9/(14))(sec^(−1) (((3x)/7))−((√(9x^2 −49))/(3x)).(7/(3x)))+c  =(9/(14))(sec^(−1) (((3x)/7))−((7(√(9x^2 −49)))/(9x^2 )))+c
(3x)272x3dx3x=7secθdx=73secθtanθdθ=(7secθ)272(73secθ)3.73secθtanθdθ=7tanθ34327sec3θ.73secθtanθdθ=97tan2θsec2θdθ=97sin2θdθ=971cos2θ2dθ=914(θ12sin2θ)=914(θsinθcosθ)+c=914(sec1(3x7)9x2493x.73x)+c=914(sec1(3x7)79x2499x2)+c
Commented by Hanuda354 last updated on 14/Oct/24
Thanks
Thanks
Answered by Ghisom last updated on 14/Oct/24
∫ ((√(9x^2 −49))/x^3 )dx=       [t=arcsin ((14(√(9x^2 −49)))/x^2 ) → dx=−((x(√(9x^2 −49)))/(14))dt]  =−(9/(28))∫(1+cos t)dt=−(9/(28))(t+sin t)=  =−((√(9x^2 −49))/(2x^2 ))−(9/(28))arcsin ((14(√(9x^2 −49)))/x^2 ) +C
9x249x3dx=[t=arcsin149x249x2dx=x9x24914dt]=928(1+cost)dt=928(t+sint)==9x2492x2928arcsin149x249x2+C
Commented by Hanuda354 last updated on 14/Oct/24
Thank you
Thankyou

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