Question Number 104851 by bemath last updated on 24/Jul/20

∫ ((√(x^2 −9))/x^3 ) dx

Answered by Dwaipayan Shikari last updated on 24/Jul/20

∫((t^2 dt)/x^4 )                                                                   x^2 −9=t^2 ⇒2x=2t(dt/dx)  ∫((t^2 dt)/((t^2 +9)^2 ))  ∫((t^2 +9)/((t^2 +9)^2 ))dt−(9/((t^2 +9)^2 ))dt  (1/3)tan^(−1) (t/3)−∫((3.9sec^2 θ)/(9(tan^2 θ+1)^2 ))dθ                  t=3tanθ  (1/3)tan^(−1) ((√(x^2 −9))/3)−3∫cos^2 θdθ  (1/3)tan^(−1) ((√(x^2 −9))/3)−((3θ)/2)−((3sin2θ)/4)  (1/3)tan^(−1) ((√(x^2 −9))/3)−((tan^(−1) (t/3))/2)−(3/4)sin(2tan^(−1) (t/3))  (1/3)tan^(−1) ((√(x^2 −9))/3)−(3/2)tan^(−1) ((√(x^2 −9))/3)−(3/4)sin(2tan^(−1) (t/3))+C

Commented by1549442205PVT last updated on 24/Jul/20

Sir mistaked and don′t detail  at this place:  −∫((9dt)/((t^2 +9)^2 ))=−∫((3.9sec^2 θdθ)/((9tan^2 θ+9)^2 ))=−(1/3)∫cos^2 θdθ  −(1/3)∫cos^2 θdθ=−(1/3)∫((1+cos2θ)/2)dθ  =((−θ)/6)−((sin2θ)/(12))=−(1/6)tan^(−1) (t/3)−(1/(12))×((2tanθ)/(1+tan^2 θ))  =−(1/6)tan^(−1) ((√(x^2 −9))/3)−(1/(12))×((2(t/3))/(1+(t^2 /9)))  =−(1/6)tan^(−1) ((√(x^2 −9))/3)−((√(x^2 −9))/(2x^2 ))  Result=(1/6)tan^(−1) ((√(x^2 −9))/3)−((√(x^2 −9))/(2x^2 ))

Answered by ~blr237~ last updated on 24/Jul/20

∫−(1/3)(√(1−((3/x))^2 ))  d((3/x))  ∫−(1/3)(√(1−u^2 )) du      u=(3/x)    ∫−(1/3)sh^2 tdt  =∫−(1/6)(ch(2t)−1)dt    u=cht  ⇒ t=ln(u−(√(u^2 −1)))  −(1/6)((1/2)sh(2t)−t)+c = −(1/6)cht(√(1−ch^2 t)) +(t/6) +c  let for you to conclude

Commented by1549442205PVT last updated on 25/Jul/20

your next way don′t lead to result  put u=sinθ⇒du=cosθdθ (u=(3/x))  −(1/3)∫(√(1−u^2 ))du=−(1/3)∫cos^2 θdθ  =−(1/3)∫((1+cos2θ)/2)dθ=((−θ)/6)−(1/(12))sin2θ  =−(1/6)sin^(−1) ((3/x))−(1/6)u(√(1−u^2 ))  =−(1/6)sin^(−1) ((3/x))−(1/(2x))(√(1−(9/x^2 )))  Result=−(1/6)sin^(−1) ((3/x))−((√(x^2 −9))/(2x^2 ))+C  you can prove that (𝛑/2)−sin^(−1) ((3/x))=  tan^(−1) (((√(x^2 −9))/3)) so this result coincide   to my answer and don′t doubt!

Answered by bramlex last updated on 24/Jul/20

∫ ((x(√(1−(9/x^2 ))))/x^3 ) dx = ∫ ((√(1−(9/x^2 )))/x^2 ) dx  ∫−(√(1−(9/x^2 ))) d((1/x)) [ let (1/x) = v ]  ∫−(√(1−9v^2 )) dv = −{((3v)/2)(√(1−9v^2 )) +(1/2)sin^(−1) (3v)}+c  =−(3/(2x))(√(1−(9/x^2 ) ))−(1/2)sin^(−1) ((3/x))+C  =−(3/2)(√(x^2 −9))−(1/2)sin^(−1) ((3/x))+C

Commented by1549442205PVT last updated on 25/Jul/20

false result because ∫_4 ^5  ((√(x^2 −9))/x^3 )dx  ≠(−(3/2)(√(x^2 −9))−(1/2)sin^(−1) ((3/x)))∣_4 ^5   Sir mistaked at this place:  ∫−(√(1−9v^2 ))dv=−v(√(1−9v^2 ))+∫  ((v.(−9v)dv)/(√(1−9v^2 )))  and this way don′t lead to final goal

Answered by 1549442205PVT last updated on 24/Jul/20

Putting (√(x^2 −9))=u⇒u^2 =x^2 −9,x=(√(u^2 +9))  ⇒udu=xdx⇒((√(x^2 −9))/x^3 )dx=((u^2 du)/((u^2 +9)^2 ))  F=∫(([(u^2 +9)−9]du)/((u^2 +9)^2 ))=∫(du/(u^2 +9))−9∫(du/((u^2 +9)^2 ))  =(1/3)tan^(−1) (u/3)−9I_2   Apply current formular:  I_n =∫(dx/((t^2 +a^2 )))=(1/(2a^2 (n−1))).(t/((t^2 +a^2 )^(n−1) ))+(1/a^2 ).((2n−3)/(2n−2)).I_(n−1)   we get  I_2 =(1/(18(2−1))).(u/((u^2 +9)^(2−1) ))+(1/9).((2.2−3)/(2.2−2)).I_1   =(u/(18(u^2 +9)))+(1/(18)).∫(du/((u^2 +9)))  =(u/(18(u^2 +9)))+(1/(18))×(1/3)tan^(−1) (u/3).Hence,  F=(1/3)tan^(−1) (u/3)−((u/(2(u^2 +9)))+(1/6)tan^(−1) (u/3))+C  =(1/6)tan^(−1) (u/3)−(u/(2(u^2 +9)))+C  F=(1/6)tan^(−1) ((√(x^2 −9))/3)−((√(x^2 −9))/(2x^2 ))+C