a) $\displaystyle\int (5x^3 - 3x + 6)dx$
$=5\displaystyle\int x^3dx - 3\displaystyle\int xdx + 6\displaystyle\int dx$
$= \dfrac{5}{4}x^4 - \dfrac32x^2 + 6x + C$
b) $\displaystyle\int(x^2 + 9x -1)dx$
$= \displaystyle\int x^2dx + 9\displaystyle\int xdx - \displaystyle\int dx$
$= \dfrac13x^3 + \dfrac92x^2 - x + C$
c) $\displaystyle\int\left(\sin2x - \dfrac{2}{\sin^2x}\right)dx$
$= \displaystyle\int\sin2xdx - 2\displaystyle\int\dfrac{1}{\sin^2x}dx$
Đặt $u = 2x$
$\to du = 2dx$
Ta được:
$\dfrac12\displaystyle\int\sin udu - 2\displaystyle\int\dfrac{1}{\sin^2x}dx$
$= -\dfrac12\cos u + 2\cot x + C$
$= 2\cot x -\dfrac12\cos2x + C$
d) $\displaystyle\int\left(e^{2x} - \dfrac{2}{3x + 1} + 4^x\right)dx$
$= \displaystyle\int e^{2x}dx - 2\displaystyle\int\dfrac{1}{3x +1}dx + \displaystyle\int4^xdx$
Đặt $u =3x + 1$
$\to du = 3dx$
Ta được:
$\displaystyle\int e^{2x}dx - \dfrac23\displaystyle\int\dfrac{du}{u} + \displaystyle\int4^xdx$
$= \dfrac12e^{2x} - \dfrac23\ln\vert u\vert + \dfrac{4^x}{\ln4} + C$
$= \dfrac{e^{2x}}{2} - \dfrac23\ln\vert3x +1\vert + \dfrac{4^x}{\ln4} + C$
e) $\displaystyle\int\left(\sqrt[3]{x^2} +\dfrac{1}{e^{3x}} + \dfrac3x\right)dx$
$= \displaystyle\int x^{\tfrac23}dx + \displaystyle\int e^{-3x}dx + 3\displaystyle\int\dfrac{dx}{x}$
$= \dfrac35x^{\tfrac53} -\dfrac13e^{-3x} + 3\ln\vert x\vert + C$
f) $\displaystyle\int\left(\cos3x +\dfrac{1}{x^3} -\sqrt{x^3}\right)dx$
$= \displaystyle\int\cos3xdx + \displaystyle\int\dfrac{1}{x^{3}}dx - \displaystyle\int\sqrt{x^3}dx$
Đặt $\begin{cases}u = 3x\\v = \sqrt x\end{cases}\to \begin{cases}du = 3dx\\ dv = \dfrac{1}{2\sqrt x}dx\end{cases}$
Ta được:
$\dfrac13\displaystyle\int\cos udu +\displaystyle\int\dfrac{1}{x^3}dx - 2\displaystyle\int v^4dv$
$= \dfrac13\sin u -\dfrac{1}{2x^2} - \dfrac25v^5 + C$
$= \dfrac13\sin3x - \dfrac{1}{2x^2} - \dfrac25\sqrt{x^5} + C$
