Question

integral of x²(x + 1)² dx​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm {x}^{2} {(x + 1)}^{2} \: dx$

$\rm \:  =  \: \displaystyle\int\rm {x}^{2}( {x}^{2} + 1 + 2x) \: dx$

$\rm \:  =  \: \displaystyle\int\rm ( {x}^{4} + {x}^{2} + 2 {x}^{3}) \: dx$

We know,

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

So, using this identity, we get

$\rm \:  =  \: \dfrac{ {x}^{4 + 1} }{4 + 1} + \dfrac{ {x}^{2 + 1} }{2 + 1} + 2 \times \dfrac{ {x}^{3 + 1} }{3 + 1} + c$

$\rm \:  =  \: \dfrac{ {x}^{5} }{5} + \dfrac{ {x}^{3} }{3} + \dfrac{ {x}^{4} }{2} + c$

Hence,

$\rm \:\boxed{\tt{ \displaystyle\int\rm {x}^{2} {(x + 1)}^{2} dx =  \: \dfrac{ {x}^{5} }{5} + \dfrac{ {x}^{3} }{3} + \dfrac{ {x}^{4} }{2} + c}}$

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Explore More

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

Answer 2

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