Short Question 1

Question 1:
$\begin{array}{l}\text{Find the integral of each of the following}\text{.}\\ (a){\displaystyle \int \left(\frac{3}{x^2}-\frac{5}{2x^3}+2\right)dx}\\ (b){\displaystyle \int x^2\left(x^5+2x\right)dx}\\ (c){\displaystyle \int \frac{3x^4+2x}{x^3}dx}\\ (d){\displaystyle \int \frac{(7+x)(7-x)}{x^4}dx}\\ (e){\displaystyle \int {(5x-1)}^{3}dx}\\ (f){\displaystyle \int \frac{3}{{\left(4x+7\right)}^8}dx}\end{array}$


Solution:
(a)
$\begin{array}{l}{\displaystyle \int \left(\frac{3}{x^2}-\frac{5}{2x^3}+2\right)dx}\\ ={\displaystyle \int \left(3x^{-2}-\frac{5x^{-3}}{2}+2\right)dx}\\ =-3x^{-1}+\frac{5x^{-2}}{4}+2x+c\\ =-\frac{3}{x}+\frac{5}{4x^2}+2x+c\end{array}$


(b)
$\begin{array}{l}{\displaystyle \int x^2\left(x^5+2x\right)dx}\\ ={\displaystyle \int \left(x^7+2x^3\right)}dx\\ =\frac{x^8}{8}+\frac{2x^4}{4}+c\\ =\frac{x^8}{8}+\frac{x^4}{2}+c\end{array}$


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


(d)
$\begin{array}{l}{\displaystyle \int \frac{(7+x)(7-x)}{x^4}dx}={\displaystyle \int \left(\frac{49-x^2}{x^4}\right)}dx\\ ={\displaystyle \int \left(\frac{49}{x^4}-\frac{1}{x^2}\right)dx}={\displaystyle \int \left(49x^{-4}-x^{-2}\right)dx}\\ =\frac{49x^{-3}}{-3}+\frac{1}{x}+c\\ =-\frac{49}{3x^3}+\frac{1}{x}+c\end{array}$


(e)
$\begin{array}{l}{\displaystyle \int {(5x-1)}^{3}dx}\\ =\frac{{(5x-1)}^4}{\left(4\right)\left(5\right)}+c\\ =\frac{1}{20}{\left(5x-1\right)}^4+c\end{array}$


(f)
$\begin{array}{l}{\displaystyle \int \frac{3}{{\left(4x+7\right)}^8}dx}={\displaystyle \int 3{\left(4x+7\right)}^{-8}dx}\\ =\frac{3{\left(4x+7\right)}^{-7}}{\left(-7\right)\left(4\right)}+c\\ =-\frac{3}{28{\left(4x+7\right)}^7}+c\end{array}$