∫(e⁶ˣ/√(9-e⁴ˣ))dx.
Let e²ˣ=3sinθ, 2e²ˣdx=3cosθdθ, dx=3cosθdθ/2e²ˣ.
√(9-e⁴ˣ)=3cosθ, e⁴ˣ=9sin²θ.
Integral becomes (9/2)∫sin²θdθ=(9/4)∫(1-cos(2θ))dθ.
Evaluate integral: (9/4)[θ-sinθcosθ]+C, because sin(2θ)=2sinθcosθ.
(9/4)[arcsin(e²ˣ/3)-(e²ˣ/9)√(9-e⁴ˣ)]+C.