∫(4dx/x2√4-9x2)=

Question

∫(4dx/x2√4-9x2)= (A) √4-9x2+C (B) (-2/3)√4-9x2+C (C) (-√4-9x2/x)+C (D) (2/3)(√4-9x2/x)+C

Tardigrade Admin · Accepted Answer

Let I=∫ (4 d x/x2 √4-9 x2)=4 ∫ (d x/x2 √(2)2-(3 x)2) put 3 x=2 sin θ ⇒ 3 d x=2 cos θ ⋅ d θ Then, I=4 × (2/3) ∫ ( cos θ d θ/(4)9 sin 2 θ √4-4 sin 2 θ) =6 ∫ ( cos θ d θ/ sin 2 θ ⋅ 2 cos θ)=3 ∫ cosec2 θ d θ =-3 cot θ+C=-3 (√1- sin 2 θ/ sin θ)+C =(-3 √1-((3 x/2))2/((3 x)2))+C =-(2/x) (√4-9 x2/2)+C =-(√4-9 x2/x)+C

Q. $\int \frac{4 d x}{x ^{2} 4 - 9 x ^{2}} =$

$4 - 9 x^{2} + C$

$\frac{- 2}{3} 4 - 9 x^{2} + C$

$\frac{- 4 - 9 x ^{2}}{x} + C$

$\frac{2}{3} \frac{4 - 9 x ^{2}}{x} + C$

$- 3 4 - 9 x^{2} + C$

Q. ∫x24−9x2​4dx​=

4−9x2​+C

3−2​4−9x2​+C

x−4−9x2​​+C

32​x4−9x2​​+C