Question

Find the following integrals:
$\left(i\right)\int \frac{(x^3-1)}{x^2}dx$
$\left(ii\right)\int (x^{\frac{2}{3}}+1)dx$
$\left(iii\right)\int (x^{\frac{3}{2}}+2e^x-\frac{1}{x})dx$

Answer 1

$\left(i\right)\int \frac{(x^3-1)}{x^2}dx=\int (\frac{x^3}{x^2}-\frac{1}{x^2})dx$
$=\int (x-\frac{1}{x^2})dx$
$=\int xdx-\int \frac{1}{x^2}dx$
$=\int x^{1}dx-\int x^{-2}.dx=\left[\left(\frac{x^{1+1}}{1+1}\right)\right]-\left[\left(\frac{x^{-2+1}}{-2+1}\right)\right]+C$
$=\frac{x^2}{2}-\frac{x^{-1}}{-1}+C$
$=\frac{x^2}{2}+\frac{1}{x}+C$
Where $C$ is constant of integration

$\left(ii\right)\int (x^{\frac{2}{3}}+1)dx$
$=\int x^{\frac{2}{3}}dx+\int dx$
$=\left[\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}\right]+x+C$
$=\frac{x^{\frac{5}{3}}}{\frac{5}{3}}+x+C$
$=\frac{3x^{\frac{5}{3}}}{5}+x+C$
Where $C$ is constant of integration

$\left(iii\right)\int (x^{\frac{3}{2}}+2e^x-\frac{1}{x})dx$
$=\int x^{\frac{3}{2}}dx+2\int e^{x}dx-\int \frac{1}{x}dx$
$=\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2e^x-\log |x|+C$
$=\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+2e^x-\log |x|+C$
$=\frac{2x^{\frac{5}{2}}}{5}+2e^x-\log |x|+C$

Where $C$ is constant of integration