1个回答
展开全部
let
(2/3)^x = sinu
ln(2/3) .(2/3)^x dx = cosu du
--------------
∫ 2^x.3^x/(9^x -4^x) dx
=∫ (2/3)^x/(1 -(4/9)^x) dx
=[1/ln(2/3)] ∫ (1/cosu) du
=[1/ln(2/3)] ∫ secu du
=[1/ln(2/3)] ln|secu+ tanu| + C
=[1/ln(2/3)] ln|1/√[1- (4/9)^x] + (2/3)^x/√[1- (4/9)^x]| + C
--------
where
(2/3)^x = sinu
cosu = √[1- (4/9)^x]
(2/3)^x = sinu
ln(2/3) .(2/3)^x dx = cosu du
--------------
∫ 2^x.3^x/(9^x -4^x) dx
=∫ (2/3)^x/(1 -(4/9)^x) dx
=[1/ln(2/3)] ∫ (1/cosu) du
=[1/ln(2/3)] ∫ secu du
=[1/ln(2/3)] ln|secu+ tanu| + C
=[1/ln(2/3)] ln|1/√[1- (4/9)^x] + (2/3)^x/√[1- (4/9)^x]| + C
--------
where
(2/3)^x = sinu
cosu = √[1- (4/9)^x]
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询