Đáp án:
$D = \dfrac{2(x-1)^7}{7} + \dfrac{(x-1)^6}{3} + C$
$A = \sqrt{x^2 - 2x + 5}+C$
Giải thích các bước giải:
$\begin{array}{l}+) \quad D = \displaystyle\int2x(x-1)^5dx\\ Đặt\,\,u = x-1\longrightarrow x = u + 1\\ \to du = dx\\ \text{Ta được:}\\ \quad D = 2\displaystyle\int(u+1)u^5du\\ \to D = 2\displaystyle\int(u^6 + u^5)du\\ \to D = 2\displaystyle\int u^6du + 2\displaystyle\int u^5du\\ \to D = \dfrac{2u^7}{7} + \dfrac{u^6}{3} + C\\ \to D = \dfrac{2(x-1)^7}{7} + \dfrac{(x-1)^6}{3} + C\\ +) \quad A = \displaystyle\int\dfrac{x-1}{\sqrt{x^2 - 2x + 5}}dx\\ Đặt\,\,u = x^2 -2x + 5\\ \to du = 2(x-1)dx\\ \text{Ta được:}\\ \quad A = \dfrac12\displaystyle\int\dfrac{1}{\sqrt u}du\\ \to A = \dfrac12\cdot 2\sqrt u + C\\ \to A = \sqrt u + C\\ \to A = \sqrt{x^2 - 2x + 5}+C \end{array}$
