Question

$\sf\red{Question:-}$


$\tt\displaystyle \tt{\int \frac{12y^{3} - 4y}{2y} dy}$

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Answer 1

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

Given integral is

$\displaystyle \tt{\int \frac{12y^{3} - 4y}{2y} dy}$

$\rm \: = \: \displaystyle \tt{\int \frac{4y(3y^{2} - 1)}{2y} dy}$

$\rm \: = \: \displaystyle\tt\int \:2( {3y}^{2} - 1) \: dy$

$\rm \: = \: 2\displaystyle\tt\int \:( {3y}^{2} - 1) \: dy$

$\rm \: = \: 6\displaystyle\tt\int \:{y}^{2} \: dy \: - \:2 \: \displaystyle\tt\int \:dy$

We know,

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

So, using this result, we get

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

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

$\rm \: = \: {2y}^{3} - 2y + c$

Hence,

$\rm\implies \:\boxed{\tt{ \: \: \rm \:\displaystyle \tt{\int \frac{12y^{3} - 4y}{2y} dy} = \: {2y}^{3} - 2y + c \: \: }}$

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ADDITIONAL INFORMATION

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

Answer 2

$\huge \color{navy}\maltese \bold \color{red} \: answer$