Question

giúp mình với ậ

Answer 1

1/ \(u=x-3\Rightarrow du=dx\Rightarrow I_3=\int\limits^4_0u^{\dfrac{1}{2}}du=\dfrac{2}{3}.u^{\dfrac{3}{2}}|^4_0=\dfrac{2}{3}.4^{\dfrac{2}{3}}=..\)

2/ \(K_3=\int\limits^1_0\dfrac{1}{\left(x-3\right)\left(x-2\right)}dx=\int\limits^1_0(\dfrac{1}{x-3}-\dfrac{1}{x-2})dx\)

\(=\int\limits^1_0\left(x-3\right)^{-1}dx-\int\limits^1_0\left(x-2\right)^{-1}dx\)

\(=\int\limits^1_0\left(x-3\right)^{-1}d\left(x-3\right)-\int\limits^1_0\left(x-2\right)^{-1}d\left(x-2\right)\)

\(=ln\left|x-3\right||^1_0-ln\left|x-2\right||^1_0=...\)