Question

Integration of —
$\ \\ \huge \int\sqrt{(3x - 4)^{3} } \: dx$

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(only for class 11th & 12th students)
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Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \sqrt{(3x - 4)^{3} } \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\sf {\bigg( {(3x - 4)}^{3} \bigg) }^{ \dfrac{1}{2} } \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\sf {\bigg( {3x - 4} \bigg) }^{ \dfrac{3}{2} } \: dx$

We know,

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf {(ax + b)}^{n} = \frac{ {(ax + b)}^{n + 1} }{(n + 1)a} + c}}}$

So, using this, we get

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

$\rm \:  =  \: \dfrac{{\bigg[3x - 4 \bigg]}^{\dfrac{3 + 2}{2}}}{\bigg[\dfrac{3 + 2}{2}\bigg] \times 3} + c$

$\rm \:  =  \: \dfrac{{\bigg[3x - 4 \bigg]}^{\dfrac{5}{2}}}{\bigg[\dfrac{5}{2}\bigg] \times 3} + c$

$\rm \:  =  \: \dfrac{2}{15} {\bigg[3x - 4 \bigg]}^{\dfrac{5}{2}} + c$

Hence,

$\red{ \boxed{ \sf{ \:\displaystyle\int\bf \sqrt{(3x - 4)^{3} } \: dx  =  \: \dfrac{2}{15} {\bigg[3x - 4 \bigg]}^{\dfrac{5}{2}} + c}}}$

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

$\sf \displaystyle \int \: (3x - 4) {}^{ \frac{3}{2} } \: dx$

$\sf = \frac{ {(3x - 4)}^{ \frac{3}{2} + 1 } }{ (\frac{3}{2} + 1) \times 3} + c$

$\sf = \frac{ {(3x - 4)}^{ \frac{5}{2} } }{ (\frac{5}{2}) \times 3} + c$

$\sf = \frac{ {(3x - 4)}^{ \frac{5}{2} } }{ (\frac{15}{2})} + c$

$\sf = \frac{2 {(3x - 4)}^{ \frac{5}{2} } }{15} + c$