Question

find the integral -
${x}^{ - 3} (x + 1)dx$
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Answer 1

Answer:

$\mathrm{-\bigg(\dfrac{1}{x} + \dfrac{1}{2x^2}\bigg) + c}$

Step-by-step explanation:

Given :

$\mathrm{{x}^{ - 3} (x + 1)dx}$

To find :

integral

Solution :

$\longmapsto \displaystyle \int \mathrm{{x}^{ - 3} (x + 1)dx}$

We know that,

$\boxed{\mathrm{a^{-m} = \dfrac{1}{a^m}}}$

$\longmapsto \displaystyle \int \mathrm{\bigg(\dfrac{x}{x^{3}} + \dfrac{1}{x^3}\bigg)dx}$

$\longmapsto \displaystyle \int \mathrm{\bigg(\dfrac{1}{x^{2}} + \dfrac{1}{x^3}\bigg)dx}$

We know that,

$\boxed{\mathrm{\int x^n = \dfrac{x^{n+1}}{n+1} + c \ ;\ n \neq 1 }}$

So,

$\implies \mathrm{\dfrac{-1}{x} + \dfrac{-1}{2x^2} + c}$

$\implies \mathrm{-\bigg(\dfrac{1}{x} + \dfrac{1}{2x^2}\bigg) + c}$

$\therefore \underline{\boxed{ \displaystyle \int \mathbf{{x}^{ - 3} (x + 1)dx} = \mathbf{-\bigg(\dfrac{1}{x} + \dfrac{1}{2x^2}\bigg) + c} }}$

❖ Extra information :

⟡ Some basic integrals :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx+c \\\\ \sf x^n \ (n \neq -1)& \sf \dfrac{x^{n+1}}{n+1} + c \\\\ \sf \dfrac{1}{x} & \sf logx+ c\\\\ \sf {e}^{x} & \sf {e}^{x}+c\\\\ \sf sinx & \sf - \: cosx+ c \\\\ \sf cosx & \sf \: sinx + c\\\\ \sf {sec}^{2} x & \sf tanx + c\\\\ \sf {cosec}^{2}x & \sf - cotx+ c \\\\ \sf secx \: tanx & \sf secx + c\\\\ \sf cosecx \: cotx& \sf -\: cosecx + c\end{array}} \\ \end{gathered}$

Hope it helps!!