g) $\displaystyle\int(4x-1)\ln(x+1)dx$
Đặt $\begin{cases}u =\ln(x+1)\\dv =(4x-1)dx\end{cases}\longrightarrow \begin{cases}du =\dfrac{1}{x+1}dx\\v =\dfrac{(4x-1)^2}{8}\end{cases}$
Ta được:
$=\dfrac18(4x-1)^2\ln(x+1) - \dfrac18\displaystyle\in\dfrac{(4x-1)^2}{x+1}dx$
$= \dfrac18(4x-1)^2\ln(x+1) -\dfrac18\displaystyle\int\left(16x +\dfrac{25}{x+1} -24\right)dx$
$= \dfrac18(4x-1)^2\ln(x+1) -2\displaystyle\int xdx - \dfrac{25}{8}\displaystyle\int\dfrac{dx}{x+1} + 3\displaystyle\int dx$
$= \dfrac18(4x-1)^2\ln(x+1) - x^2 -\dfrac{25}{8}\ln|x+1| + 3x + C$
h) $\displaystyle\int x\sin^2xdx$
$= \dfrac12\displaystyle\int x(1-\cos2x)dx$
$= \dfrac12\displaystyle\int xdx - \dfrac12\displaystyle\int x\cos2xdx$
Đặt $\begin{cases}u = x\\dv = \cos2xdx\end{cases}\longrightarrow \begin{cases}du = dx\\v =\dfrac12\sin2x\end{cases}$
Ta được:
$\dfrac12\displaystyle\int xdx - \dfrac12\left(\dfrac{\sin2x}{2} -\dfrac12\displaystyle\int\sin2xdx\right)$
$= \dfrac14x^2 - \dfrac14\sin2x -\dfrac14\cdot\dfrac{\cos2x}{2} + C$
$=\dfrac{x^2}{4} -\dfrac{\sin2x}{4} - \dfrac{\cos2x}{8} + C$
