Question

$$(i)\int\frac{ix}{x+2}dx$$

Answer

$$-(e^2*IM*n*t^2*g*r*a*x^2*d)/(x+2)$$

Solution


Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{\imath \imath ntegrate\imath xdx}{x+2}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{{\imath }^{3}n{t}^{2}{e}^{2}gra{x}^{2}d}{x+2}\]
Isolate \({\imath }^{2}\).
\[\frac{{\imath }^{2}\imath n{t}^{2}{e}^{2}gra{x}^{2}d}{x+2}\]
Use Square Rule: \({i}^{2}=-1\).
\[\frac{-1\times \imath n{t}^{2}{e}^{2}gra{x}^{2}d}{x+2}\]
Simplify  \(1\times \imath n{t}^{2}{e}^{2}gra{x}^{2}d\)  to  \(n{t}^{2}gra{x}^{2}d\imath {e}^{2}\).
\[\frac{-n{t}^{2}gra{x}^{2}d\imath {e}^{2}}{x+2}\]
Regroup terms.
\[\frac{-{e}^{2}\imath n{t}^{2}gra{x}^{2}d}{x+2}\]
Move the negative sign to the left.
\[-\frac{{e}^{2}\imath n{t}^{2}gra{x}^{2}d}{x+2}\]