Question

$\huge{ \boxed {\bold \red{{ \int( \frac{1}{ {x}^{2} } + \frac{6}{ {x}^{3}})dx}}}}$



❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​

Answer 1

$\bold \red{\int{\left(\frac{1}{x^{2}} + \frac{6}{x^{3}}\right)d x}}$

Integrate term by term:

${ \bold \red{\color{red}{\int{\left(\frac{1}{x^{2}} + \frac{6}{x^{3}}\right)d x}} = \color{red}{\left(\int{\frac{6}{x^{3}} d x} + \int{\frac{1}{x^{2}} d x}\right)}}}$

Apply the power rule

${ \bold \red{ \int x^{n}\, dx = \frac{x^{n + 1}}{n + 1} \left(n \neq -1 \right) \: with \: n=-2:}}$

${\bold \red{\int{\frac{6}{x^{3}} d x} + \color{red}{\int{\frac{1}{x^{2}} d x}}=\int{\frac{6}{x^{3}} d x} + \color{red}{\int{x^{-2} d x}}=\int{\frac{6}{x^{3}} d x} + \color{red}{\frac{x^{-2 + 1}}{-2 + 1}}=\int{\frac{6}{x^{3}} d x} + \color{red}{\left(- x^{-1}\right)}=\int{\frac{6}{x^{3}} d x} + \color{red}{\left(- \frac{1}{x}\right)}}}$

Apply the constant multiple rule

${ \bold \red{\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx \: with \: c=6 \: and \: f{\left(x \right)} = \frac{1}{x^{3}}:}}$

$\bold \red{\color{red}{\int{\frac{6}{x^{3}} d x}} - \frac{1}{x} = \color{red}{\left(6 \int{\frac{1}{x^{3}} d x}\right)} - \frac{1}{x}}$

Apply the power rule

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

${\bold \red{6 \color{red}{\int{\frac{1}{x^{3}} d x}} - \frac{1}{x}=6 \color{red}{\int{x^{-3} d x}} - \frac{1}{x}=6 \color{red}{\frac{x^{-3 + 1}}{-3 + 1}} - \frac{1}{x}=6 \color{red}{\left(- \frac{x^{-2}}{2}\right)} - \frac{1}{x}=6 \color{red}{\left(- \frac{1}{2 x^{2}}\right)} - \frac{1}{x}}}$

Therefore,

$\bold \red{\int{\left(\frac{1}{x^{2}} + \frac{6}{x^{3}}\right)d x} = - \frac{1}{x} - \frac{3}{x^{2}}}$

Simplify:

$\bold \red{\int{\left(\frac{1}{x^{2}} + \frac{6}{x^{3}}\right)d x} = - \frac{x + 3}{x^{2}}}$

Add the constant of integration:

$\bold \red{\int{\left(\frac{1}{x^{2}} + \frac{6}{x^{3}}\right)d x} = - \frac{x + 3}{x^{2}}+C}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\red{\rm :\longmapsto\:\displaystyle\int\sf \bigg[\dfrac{1}{ {x}^{2} } + \dfrac{6}{ {x}^{3} } \bigg] \: dx \: }$

can be rewritten as

$\rm \: = \: \displaystyle\int\sf \frac{1}{ {x}^{2} } \: dx \: + \: \displaystyle\int\sf \frac{6}{ {x}^{3} } \: dx$

$\rm \: = \: \displaystyle\int\sf {x}^{ - 2} \: dx \: + \: 6 \: \displaystyle\int\sf {x}^{ - 3} \: dx$

We know

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

So, using this result, we get

$\rm \: = \: \dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} + 6 \times \dfrac{ {x}^{ - 3 + 1} }{ - 3 + 1} \: + \: c$

$\rm \: = \: \dfrac{ {x}^{ - 1} }{ - 1} + 6 \times \dfrac{ {x}^{ - 2} }{ - 2} \: + \: c$

$\rm \: = \: - \: \dfrac{1}{x} - \dfrac{3}{ {x}^{2} } + c$

$\rm \: = \: - \:\bigg( \dfrac{1}{x} + \dfrac{3}{ {x}^{2} }\bigg) + c$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\int\sf \bigg[\dfrac{1}{ {x}^{2} } + \dfrac{6}{ {x}^{3} } \bigg] \: dx \: = \: - \:\bigg( \dfrac{1}{x} + \dfrac{3}{ {x}^{2} }\bigg) + c}}}$

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Additional Information :-

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