Question

PLEASE SOLVE MY QUESTION!!​

Answer 1

Step-by-step explanation:

We have,

$\sf{ I = \int {x}^{ - 3} \: dx }$

$\sf{ \implies \: I = \dfrac{ {x}^{ - 3 + 1} }{ - 3 + 1} + C }$

$\sf{ \implies \: I = \dfrac{ {x}^{ - 2} }{ - 2} + C }$

$\sf{ \implies \: I = \dfrac{1}{ - 2 {x}^{2} } + C }$

$\sf{ \implies \: I = - \dfrac{1}{ 2 {x}^{2} } + C }$

Answer 2

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \: \to \: given \: : \\ \\ I \: = \displaystyle \int \: \rm \: {x}^{ - 3} \: dx\\ \\ \underbrace{{\pink{ { \boxed{ \begin{array}{cc} \sf \: we \: know \: that \: : \\ \\ \sf \hookrightarrow \: \displaystyle \int \: \rm \: {x}^{n} \: dx = \frac{ {x}^{n + 1} }{ n+ 1} + c\end{array}}}}}}_{ \underbrace{ \sf \: apply \: this \: rule}_ { \downarrow}} \\ \\ \rm = \frac{ {x}^{ - 3 + 1} }{ - 3 + 1} + c \\ \\ \rm = \frac{ {x}^{ - 2} }{ - 2} + c \\ \\ \rm = - \frac{ { {(x}^{ 2} })^{ - 1} }{ 2} + c \\ \\ \rm = - \frac{ \frac{1}{ {x}^{2} } }{2} + c \\ \\ \rm = - \frac{1}{ {x}^{2} } \times \frac{1}{2} + c \\ \\ \rm = - \frac{1}{2 {x}^{2} } + c \end{array}}}}}$