Tìm nguyên hàm của hàm số sau: ∫12x/16x−9xdx

Question

\(\int \frac{{\rm 12}^{x} }{{\rm 16}^{x} -{\rm 9}^{x} } {\rm d}x\)

Answer 1

Ta có: \(\int \frac{{\rm 12}^{x} }{{\rm 16}^{x} -{\rm 9}^{x} } {\rm d}x =\int \frac{\left(\frac{12}{9} \right)^{x} {\rm d}x}{\left(\frac{16}{9} \right)^{x} -1}  =\int \frac{\left(\frac{4}{3} \right)^{x} {\rm d}x}{\left(\frac{4}{3} \right)^{2x} -1}  \).

Đặt \(t=\left(\frac{4}{3} \right)^{x}  \Rightarrow {\rm d}t=\left(\frac{4}{3} \right)^{x} \ln \frac{4}{3} {\rm d}x.\)

Khi đó: \(\int \frac{{\rm 12}^{x} }{{\rm 16}^{x} -{\rm 9}^{x} } {\rm d}x =\frac{1}{\ln \frac{4}{3} } \int \frac{{\rm d}t}{t^{2} -1}  =\frac{1}{2\ln \frac{4}{3} } \ln \left|\frac{t-1}{t+1} \right|+C=\frac{1}{2\ln \frac{4}{3} } \ln \left|\frac{4^{x} -3^{x} }{4^{x} +3^{x} } \right|+C.\)