# Integration, Short Questions (Question 2 – 4)

Question 2:
Given that  $\text{Given that}\int \frac{4}{{\left(1+x\right)}^{4}}dx=m{\left(1+x\right)}^{n}+c,$
find the values of m and n.

Solution:
$\begin{array}{l}\int \frac{4}{{\left(1+x\right)}^{4}}dx=m{\left(1+x\right)}^{n}+c\\ \int 4{\left(1+x\right)}^{-4}dx=m{\left(1+x\right)}^{n}+c\\ \frac{4{\left(1+x\right)}^{-3}}{-3\left(1\right)}+c=m{\left(1+x\right)}^{n}+c\\ -\frac{4}{3}{\left(1+x\right)}^{-3}+c=m{\left(1+x\right)}^{n}+c\\ m=-\frac{4}{3},\text{}n=-3\end{array}$

Question 3:

Solution:
$\begin{array}{l}{\int }_{-1}^{2}\left[mx+3g\left(x\right)\right]dx=15\\ {\int }_{-1}^{2}mxdx+{\int }_{-1}^{2}3g\left(x\right)dx=15\\ {\left[\frac{m{x}^{2}}{2}\right]}_{-1}^{2}+3{\int }_{-1}^{2}g\left(x\right)dx=15\\ \left[\frac{m{\left(2\right)}^{2}}{2}-\frac{m{\left(-1\right)}^{2}}{2}\right]+\frac{3}{2}{\int }_{-1}^{2}2g\left(x\right)dx=15\\ 2m-\frac{1}{2}m+\frac{3}{2}\left(4\right)=15←\overline{)\text{given}{\int }_{-1}^{2}2g\left(x\right)dx=4}\\ \frac{3}{2}m+6=15\\ \frac{3}{2}m=9\\ m=9×\frac{2}{3}\\ m=6\end{array}$

Question 4:
$\text{Given}\frac{d}{dx}\left(\frac{2x}{3-x}\right)=g\left(x\right)\text{, find}{\int }_{1}^{2}g\left(x\right)dx.$

Solution:
$\begin{array}{l}\text{Given}\frac{d}{dx}\left(\frac{2x}{3-x}\right)=g\left(x\right)\\ \int g\left(x\right)dx=\frac{2x}{3-x}\\ \text{Thus,}\\ {\int }_{1}^{2}g\left(x\right)dx={\left[\frac{2x}{3-x}\right]}_{1}^{2}\\ \text{}=\frac{2\left(2\right)}{3-2}-\frac{2\left(1\right)}{3-1}\\ \text{}=4-1\\ \text{}=3\end{array}$