Answer:
Hi Army!
See the attachment I hope it helps you
[tex]\qquad\boxed{ \sf{ \: \: \sf \: (1). \: \: \: \: 6x + 5log |x| + \dfrac{6}{x} + c \: \: }}\\ \\ [/tex]
Step-by-step explanation:
Given integral is
[tex]\sf \: \displaystyle\int\sf \dfrac{(2x + 3)(3x - 2)}{ {x}^{2} } \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \dfrac{ {6x}^{2} - 4x + 9x - 6 }{ {x}^{2} } \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \dfrac{ {6x}^{2} + 5x - 6 }{ {x}^{2} } \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf\left( \dfrac{ {6x}^{2} }{ {x}^{2} } + \dfrac{5x}{ {x}^{2} } - \dfrac{6}{ {x}^{2} } \right) \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf\left( 6 + \dfrac{5}{ {x}} - {6x}^{ - 2} \right) \: dx \\ \\ [/tex]
[tex]\sf \: = \: 6x + 5log |x| - 6 \times \dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} + c \\ \\ [/tex]
[tex]\sf \: = \: 6x + 5log |x| - 6 \times \dfrac{ {x}^{ -1} }{ - 1} + c \\ \\ [/tex]
[tex]\sf \: = \: 6x + 5log |x| + \dfrac{6}{x} + c \\ \\ [/tex]
Hence,
[tex]\sf\implies \: \displaystyle\int\bf \dfrac{(2x + 3)(3x - 2)}{ {x}^{2} }dx = 6x + 5log |x| + \dfrac{6}{x} + c\\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
ADDITIONAL INFORMATION
[tex]\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}[/tex]
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Answers & Comments
Answer:
Hi Army!
See the attachment I hope it helps you
Verified answer
Answer:
[tex]\qquad\boxed{ \sf{ \: \: \sf \: (1). \: \: \: \: 6x + 5log |x| + \dfrac{6}{x} + c \: \: }}\\ \\ [/tex]
Step-by-step explanation:
Given integral is
[tex]\sf \: \displaystyle\int\sf \dfrac{(2x + 3)(3x - 2)}{ {x}^{2} } \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \dfrac{ {6x}^{2} - 4x + 9x - 6 }{ {x}^{2} } \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \dfrac{ {6x}^{2} + 5x - 6 }{ {x}^{2} } \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf\left( \dfrac{ {6x}^{2} }{ {x}^{2} } + \dfrac{5x}{ {x}^{2} } - \dfrac{6}{ {x}^{2} } \right) \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf\left( 6 + \dfrac{5}{ {x}} - {6x}^{ - 2} \right) \: dx \\ \\ [/tex]
[tex]\sf \: = \: 6x + 5log |x| - 6 \times \dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} + c \\ \\ [/tex]
[tex]\sf \: = \: 6x + 5log |x| - 6 \times \dfrac{ {x}^{ -1} }{ - 1} + c \\ \\ [/tex]
[tex]\sf \: = \: 6x + 5log |x| + \dfrac{6}{x} + c \\ \\ [/tex]
Hence,
[tex]\sf\implies \: \displaystyle\int\bf \dfrac{(2x + 3)(3x - 2)}{ {x}^{2} }dx = 6x + 5log |x| + \dfrac{6}{x} + c\\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
ADDITIONAL INFORMATION
[tex]\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}[/tex]