Question

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




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

Answer 1

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

Integrate term by term:

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

The integral of

$\bold \red{\frac{1}{x} \: is \int{\frac{1}{x} d x} = \ln{\left(x \right)}}$

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

Expand the expression:

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

Integrate term by term:

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

Apply the constant multiple rule

$\bold{\red{\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx \: with \: c=6 \: and \: f{\left(x \right)} = \frac{1}{x^{2}}}}$

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

Apply the power rule

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

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

Apply the constant multiple rule

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

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

The integral of

$\bold \red{\frac{1}{x} \: is \int{\frac{1}{x} d x} = \ln{\left(x \right)}}$

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

Therefore,

$\bold \red{\int{\left(\frac{1}{x} + \frac{3 x - 6}{x^{2}}\right)d x} = 4 \ln{\left(\left|{x}\right| \right)} + \frac{6}{x}}$

Add the constant of integration:

$\bold{ \red{\int{\left(\frac{1}{x} + \frac{3 x - 6}{x^{2}}\right)d x} = 4 \ln{\left(\left|{x}\right| \right)} + \frac{6}{x}+C}}$

Answer 2

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

Given integral is

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

can be rewritten as

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

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

$\rm \: = \:\displaystyle\int\sf \bigg[ \frac{4}{x}- \frac{6}{ {x}^{2} } \bigg] \: dx$

$\rm \: = \: 4\displaystyle\int\sf \frac{1}{x} \: dx \: - 6\displaystyle\int\sf {x}^{ - 2} \: dx$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\int\sf \frac{1}{x} \: dx \: = \: logx \: + \: c \: }}}$

and

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

So using these results, we get

$\rm \: = \: 4 \: log |x| - 6 \times \dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} + c$

$\rm \: = \: 4 \: log |x| - 6 \times \dfrac{ {x}^{ - 1} }{ - 1} + c$

$\rm \: = \: 4 \: log |x| + \dfrac{ 6}{x} + c$

Hence,

$\red{\boxed{ \tt{ \: \displaystyle\int\sf \bigg[ \frac{1}{x} + \dfrac{3x - 6}{ {x}^{2} } \bigg] \: dx = \: 4 \: log |x| + \dfrac{ 6}{x} + c \: }}}$

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Additional Information :-

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