Question

integrate...................

Answer 1

We will use one identity:

$\displaystyle \boxed{\int x^n \, dx = \frac{x^{n+1}}{n+1}+c}$

Here, the problem is then solved as follows:

$\displaystyle I = \int 2x(1-x^{-3}) \, dx \\ \\ \\ \implies I = \int (2x-2x \times x^{-3}\, dx \\ \\ \\ \implies I = \int 2x \, dx - 2\int x^{-2} \, dx \\ \\ \\ \implies I = 2 \frac{x^{1+1}}{1+1} -2 \frac{x^{-2+1}}{-2+1} + c \\ \\ \\ \implies I = 2 \frac{x^2}{2} - 2 \times \frac{x^{-1}}{-1} + c \\ \\ \\ \implies \boxed{I = x^2 + \frac{2}{x}+c}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\boxed{\sf{I=\int 2x\left(1-x^{-3}\right)\:dx}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{I=\int 2x\left(1-x^{-3}\right)\:dx=x^2+\dfrac{2}{x}+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=2\cdot \int \:x\left(1-x^{-3}\right)dx$

$\text { Expand } x\left(1-x^{-3}\right): \quad x-\dfrac{1}{x^{2}}$

$=2\cdot \int \:x-\dfrac{1}{x^2}dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=2\left(\int \:xdx-\int \dfrac{1}{x^2}dx\right)$

$\begin{array}{l}\int x d x=\dfrac{x^{2}}{2} \\\\\\\int \dfrac{1}{x^{2}} d x=-\dfrac{1}{x}\end{array}$

$=2\left(\dfrac{x^2}{2}-\left(-\dfrac{1}{x}\right)\right)$

$\text { Simplify } 2\left(\dfrac{x^{2}}{2}-\left(-\dfrac{1}{x}\right)\right): x^{2}+\dfrac{2}{x}$

$=x^2+\dfrac{2}{x}$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$\boxed{\sf{=x^2+\dfrac{2}{x}+C}}$