Question

Integral
${ \boxed{ \bold \red{ \int(2x + 5) {}^{3} dx}}}$
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Answer 1

$\bold{\int {(2x+5)}^{3} \, dx}$

1. Expand

${\boxed{ \bold{\int 8{x}^{3}+60{x}^{2}+150x+125 \, dx}}}$

$\bold{2. \: Use \: Power \: Rule: \int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C}$

${\boxed{\bold{2{x}^{4}+20{x}^{3}+75{x}^{2}+125x}}}$

3. Add constant.

${\boxed{\bold{2{x}^{4}+20{x}^{3}+75{x}^{2}+125x+C}}}$

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm {(2x + 5)}^{3} \: dx$

To solve this integral, we use method of Substitution

So, Substitute

$\red{\rm :\longmapsto\:2x + 5 = y}$

$\red{\rm :\longmapsto\:2dx= dy}$

$\red{\rm :\longmapsto\:dx= \dfrac{dy}{2} }$

So, on substituting these values, we get

$\rm \: = \: \displaystyle\int\rm {y}^{3} \: \dfrac{dy}{2}$

$\rm \: = \: \dfrac{1}{2}\displaystyle\int\rm {y}^{3} \: dy$

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

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

$\rm \: = \: \dfrac{ {y}^{4} }{8} + c$

$\rm \: = \: \dfrac{ {(2x + 5)}^{4} }{8} + c$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \displaystyle\int\rm {(2x + 5)}^{3}dx = \: \dfrac{ {(2x + 5)}^{4} }{8} + c \: }}$

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Additional Information :-

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