Integrals

$\begin{array}{l}\int x^2-\frac{x^2}{x^2}dx\\ ={\displaystyle \int x^2}–1dx={\displaystyle \int x^2}dx–{\displaystyle \int 1dx}\end{array}$

$=\frac{x^3}{3}-x+C$

= $\begin{array}{l} {\displaystyle \int 4e^{3x}dx}+{\displaystyle \int 1dx}\\ \frac{4·e^{3x}}{3}+x+C\end{array}$

$\frac{d}{dx}\left(-\frac{1}{2}cos2x-\frac{4}{3}{e}^{3x}\right)=sin2x-4e^{3x}$

Therefore, an anti-derivative of $\left(sin2x-4e^{3x}\right)is-\frac{1}{2}cos2x-\frac{4}{3}{e}^{3x}$

$\frac{dx}{}{(ax+b)}^3=3a{(ax+b)}^2$

${\left(ax+b\right)}^2=\frac{1}{3a}.\frac{a}{dx}{(ax+b)}^3$

${\left(ax+b\right)}^2=\frac{d}{dx}\left(\frac{1}{3a}{(ax+b)}^3\right)$

Therefore, an anti-derivative of ${(ax+b)}^{3}is\frac{1}{3a}(ax+b){}^3$

$\begin{array}{l}\frac{d}{dx}\left(e^{2}x\right)=2e^{2x}\\ e^{2x}=\frac{1}{2}-\frac{d}{dx}\left(e^{2x}\right)\\ e^{2x}=\frac{d}{dx}\left(\frac{1}{2}{e}^{2x}\right)\end{array}$

Therefore, an anti-derivative of $e^{2x}is\frac{1}{2}{e}^{2x}$

$\begin{array}{l}\frac{dx}{}sin3x=3cos3x \\ cos3x=\frac{1}{3}dx\left(sin3x\right) \\ cos3x= \frac{d}{dx}\left(\frac{1}{3}sin3x\right)\end{array}$

Therefore, an anti-derivative of $cos3xis\frac{1}{3}sin3x$

$\begin{array}{l}\frac{d}{dx}cos2x=–2sin2x \\ sin2x=–\frac{1}{2}\frac{d}{dx}\left(cos2x\right) \\ sin2x =\frac{d}{dx}\left(\frac{1}{2}cos2x\right)\end{array}$

Therefore, an anti-derivative of $sin2xis-\frac{1}{2}cos2x$