21) $\displaystyle\int x^3\sqrt{5 + x^2}dx$
Đặt $u = x^2 + 5$
$\to du = 2xdx$
Ta được:
$=\dfrac12\displaystyle\int(u-5)\sqrt udu$
$=\dfrac12\left(\displaystyle\int \sqrt{u^3}du - 5\displaystyle\int\sqrt udu\right)$
$=\dfrac12\displaystyle\int u^{\tfrac32}du - \dfrac52\displaystyle\int \sqrt udu$
$=\dfrac12\cdot\dfrac{2u^{\tfrac52}}{5} - \dfrac52\cdot\dfrac{2u^{\tfrac32}}{3} + C$
$= \dfrac{(x^2 + 5)^{\tfrac52}}{5} - \dfrac{(x^2 + 5)^{\tfrac32}}{3} + C$
24) $\displaystyle\int x(x+1)^{2020}dx$
Đặt $u = x + 1 \longrightarrow x = u -1$
$\to du = dx$
Ta được:
$\quad \displaystyle\int(u -1)u^{2020}du$
$= \displaystyle\int u^{2021}du- \displaystyle\int u^{2020}du$
$= \dfrac{u^{2022}}{2022} -\dfrac{u^{2021}}{2021} + C$
$= \dfrac{(x+1)^{2022}}{2022} -\dfrac{(x+1)^{2021}}{2021} + C$
27) $\displaystyle\int\sin^3x.\cos^4xdx$
$= -\dfrac{\sin^2x\cos^5x}{7} + \dfrac27\displaystyle\int\sin x.\cos^4xdx$
Đặt $u = \cos x$
$\to du =-\sin xdx$
Ta được:
$ -\dfrac{\sin^2x\cos^5x}{7} -\dfrac27\displaystyle\int u^4du$
$= -\dfrac{\sin^2x\cos^5x}{7} -\dfrac27\cdot\dfrac{u^5}{5} + C$
$= -\dfrac{\sin^2x\cos^5x}{7} -\dfrac{2\cos^5x}{35} + C$
