Tích phân

Question

$\int\limits_{1}^{9}\frac{ln(16-x)}{\sqrt{x}}dx$

score 0 · Answer 1

$I = \int\limits_1^9 {\frac{{ln(16 - x)}}{{\sqrt x }}} dx$

Đặt:

$u = ln(16 - x) \Rightarrow du = \frac{{dx}}{{x - 16}}$

$dv = \frac{{dx}}{{\sqrt x }} \Leftarrow v = 2\sqrt x$

$\Rightarrow I = 2\sqrt x ln(16 - x) - 2\int\limits_1^9 {\frac{{\sqrt x }}{{x - 16}}} dx = 6\ln 7 - 2\ln 15 - 2\int\limits_1^9 {\frac{{\sqrt x }}{{x - 16}}} dx$

$K = \int\limits_1^9 {\frac{{\sqrt x }}{{x - 16}}} dx = \int\limits_1^9 {\frac{{x - 16 + 16}}{{\sqrt x (x - 16)}}} dx = \int\limits_1^9 {\frac{1}{{\sqrt x }}} dx + \int\limits_1^9 {\frac{{16}}{{\sqrt x (x - 16)}}} dx$