Class 12th

$\begin{array}{l}\int \frac{x^3-x^2}{x-1}+\frac{x-1}{x-1}dx\\ \int \left(x^2\frac{(x-1)}{x-1}+1\right)dx\\ \begin{array}{ll}={\displaystyle \int \left(x^2+1\right)}dx\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int x^2}.dx+{\displaystyle \int 1}.dx\hfill \end{array}\\ =\frac{x^3}{3}+x+\; C\end{array}$

$\begin{array}{l}={\displaystyle \int \left(x^3.x^{–1/2}+3x.x^{–1/2}+4.x^{–1/2}\right)}dx\\ =\int \left(x^{5/2}+3x^{1/2}+4x^{-1/2}\right)dx\\ =\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}+3·\frac{x^{1/2+1}}{\frac{1}{2}+1}+4·\frac{x^{-1/2+1}}{-\frac{1}{2}+1}\\ =\frac{x^{7/2}}{7/2}+\frac{3·x^{3/2}}{3/2}+\frac{4x^{1/2}}{1/2}+C\\ =\frac{2}{7}{x}^{7/2}+2x^{3/2}+8x^{1/2}+C\\ =\frac{2}{7}{x}^{7/2}+2x^{3/2}+\mathrm{8\surd x}+C.\end{array}$

$\begin{array}{l}\int \left(\frac{x^3}{x^2}+\frac{5x^2}{x^2}-\frac{4}{x^2}\right)dx\\ \begin{array}{ll}={\displaystyle \int (x+5-4x^{–2})}dx\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int x}.dx+{\displaystyle \int 5}dx–{\displaystyle \int 4x^{–2}}dx\hfill \end{array}\\ =\frac{x^2}{2}+5x-4\frac{x^{-2+1}}{-2+1}\\ =\frac{x^2}{2}+5x-\frac{4x^{-1}}{-1}\\ =\frac{x^2}{2}+5x+\frac{4}{x}+\; C\end{array}$

Kindly go through the solution

${\displaystyle \int 2x^2+e^x}dx$

$\begin{array}{l}=2·{\displaystyle \int x^2}dx+{\displaystyle \int e^x}dx\\ =\frac{2·x^3}{3}+e^x+\; C\end{array}$

$\begin{array}{l}=a.{\displaystyle \int x^2.}dx+b{\displaystyle \int x}.dx+c{\displaystyle \int 1.dx}\\ =\frac{a·x^3}{3}+b\frac{x^2}{2}+cx+d.\end{array}$

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

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

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

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

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 ${(ax+b)}^{3}is\frac{1}{3a}(ax+b){}^3$