Question

Integral dx/(4x^2 + 9)
Bisa tolong darimana dapat jawabannya 1/6 tan^-1 (2x/3) + c?

Answer 1

Ingat!
∫ 1/(x² + a²) = 1/a arc tan (x/a)

∫ dx/(4x^2 + 9) = ∫ dx/((2x)^2 + 3^2)

misal, u = 2x
du = 2 dx
dx = du/2

∫ dx/((2x)^2 + 3^2) = ∫ du/2(u^2 + 3^2)
= (1/2)∫ du/(u^2 + 3^2)
= (1/2)(1/3) arc tan (u/3) + c
= 1/6 arc tan (2x/3) + c