Question

换元积分法这个题怎么做

换元积分法这个题怎么做第三题

Answer 1

let
(2/3)^x = sinu
ln(2/3) .(2/3)^x dx = cosu du
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∫ 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]