Đáp án:
$\begin{array}{l}1)\quad -\dfrac{1}{15}(1-x^3)^5 + C\\2)\quad \ln|1 + x^3| + C\\3)\quad \ln|x^2 + x + 7| + C\\4)\quad \dfrac14\ln|1 + 2\sin2x| + C\\5)\quad \ln|1+\sin x| + C\end{array}$
Giải thích các bước giải:
$\begin{array}{l}1)\quad \displaystyle\int x^2(1-x^3)^4dx\\ Đặt\,\,u=1 - x^3\\ \to du = -3x^2dx\\ \text{Ta được:}\\ \quad -\dfrac13\displaystyle\int u^4du\\ = -\dfrac13\cdot\dfrac{u^5}{5} + C\\ = -\dfrac{1}{15}u^5 + C\\ = -\dfrac{1}{15}(1-x^3)^5 + C\\ 2)\quad \displaystyle\int\dfrac{3x^2}{1+x^3}dx\\ Đặt\,\,u = 1 + x^3\\ \to du = 3x^2dx\\ \text{Ta được:}\\ \quad \displaystyle\int\dfrac{du}{u}\\ = \ln|u| + C\\ = \ln|1 + x^3| + C\\ 3)\quad \displaystyle\int\dfrac{2x+1}{x^2 +x+7}dx\\ Đặt\,\,u = x^2 + x + 7\\ \to du = 2x + 1\\ \text{Ta được:}\\ \quad \displaystyle\int\dfrac{du}{u}\\ = \ln|u| + C\\ = \ln|x^2 + x + 7| + C\\ 4)\quad \displaystyle\int\dfrac{\cos2x}{1+\sin2x}dx\\ Đặt\,\,u = 1 + 2\sin2x\ \to du =4\cos2xdx\\ \text{Ta được:}\\ \quad \dfrac14\displaystyle\int\dfrac{du}{u}\\ = \dfrac14\cdot\ln|u| +C\\ = \dfrac14\ln|1 + 2\sin2x| + C\\ 5)\quad \displaystyle\int\dfrac{\cos x}{1+\sin x}dx\\ Đặt\,\,u = 1+ \sin x\\ \to du = \cos xdx\\ \text{Ta được:}\\ \quad \displaystyle\int\dfrac{du}{u}\\ = \ln|u| +C\\ = \ln|1+\sin x| + C \end{array}$
