Maths Integrals

$\begin{array}{l}={\displaystyle \int 2x^2}.dx–{\displaystyle \int 3sinx}.dx+{\displaystyle \int \mathrm{5\surd x}}dx\\ =\frac{2x^3}{3}+3cosx+\frac{5x^{3/2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; C\\ =\frac{2x^3}{3}+3cosx+\frac{10x^{3/2}}{3}+\; C\end{array}$

$={\displaystyle \int 2x}dx–{\displaystyle \int 3cosx}+e^{x}dx$

$\begin{array}{l}=\frac{2x^2}{2}-3sinx+e^x+\; C\\ =x^2-3sinx+e^x+C\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(3x^2.x^{1/2}+2x.x^{1/2}+3x^{1/2}\right)}.dz\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(3x^{5/2}+2x^{3/2}+3x^{4/2}\right)}.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int 3x^{5/2}}.dx+{\displaystyle \int 2x^{3/2}}.dx+{\displaystyle \int 3x^{1/2}}.dx\hfill \end{array}\\ =3·\frac{x^{7/2}}{7/2}+\frac{2·x^{5/2}}{5/2}+\frac{3·x^{3/2}}{3/2}+\; C\\ =\frac{6·x^{7/2}}{7}+\frac{4x^{5/2}}{5}+\frac{6}{3}{x}^{3/2}+\; C\\ =\frac{6x^{7/2}}{7}+\frac{4·x^{5/2}}{5}+2x^{3/2}+\; C.\end{array}$

$\begin{array}{l}\int (\surd x-x·x^{1/2})dx\\ \int (\surd x-x^{3/2})·dx\\ \int x^{1/2}·dx-\int x^{3/2}·dx\\ \frac{x^{3/2}}{3/2}-\frac{x^{5/2}}{5/2}+C\\ \frac{2}{3}{x}^{3/2}-\frac{2}{5}{x}^{5/2}+\end{array}$ c

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

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

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

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

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

Kindly go through the solution

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

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

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