Question

Integrate $\bf x^{-3}$ w.r.t.x



​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given integral.

$\displaystyle \rm =\int {x}^{ - 3} \: dx$

We know that:

$\displaystyle \rm \longrightarrow\int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} +C$

Therefore:

$\displaystyle \rm = \dfrac{ {x}^{ - 3 + 1} }{ - 3 + 1} +C$

$\displaystyle \rm = \dfrac{ {x}^{ - 2} }{ - 2} +C$

$\displaystyle \rm = \dfrac{ - 1}{2 {x}^{2} } +C$

Therefore:

$\displaystyle \rm \longrightarrow\int {x}^{ - 3} \: dx = \dfrac{ - 1}{2 {x}^{2} } +C$

$\textsf{\large{\underline{Learn More}:}}$

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$

Answer 2

$\displaystyle\sf \red{ \int{x^{-3} \: dx}}$

Can be written as

$\displaystyle\sf \red{ \int{\frac{1}{x^{3}} d x}}$

We know that:-

$\displaystyle \sf \red{\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1} \left(n \neq -1 \right) with \: n=-3}$

$\displaystyle \sf{\color{red}{\int{\frac{1}{x^{3}} d x}}=\color{red}{\int{x^{-3} d x}}=\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}=\color{red}{\left(- \frac{x^{-2}}{2}\right)}=\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$

Therefore,

$\displaystyle \sf \red{\int{\frac{1}{x^{3}} d x} = - \frac{1}{2 x^{2}}}$

$\displaystyle \sf \red{\int{\frac{1}{x^{3}} d x} = - \frac{1}{2 x^{2}}+C}$