Question

Integrate the function

$\int \: \frac{dx}{ {(x - 1)}^{ \frac{3}{4}} {(x + 2)}^{ \frac{5}{4} } }$
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Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{{\bigg(x - 1 \bigg) }^{\dfrac{3}{4}}{\bigg(x + 2\bigg) }^{\dfrac{5}{4} }}$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{dx}{{\bigg(x - 1 \bigg) }^{\dfrac{3}{4}}{\bigg(x + 2\bigg) }^{\dfrac{5}{4}} \times {\bigg(x + 2 \bigg) }^{\dfrac{3}{4}} \times \dfrac{1}{{\bigg(x + 2\bigg) }^{\dfrac{3}{4} }} }$

can be further rewritten as

$\rm \: = \: \dfrac{dx}{ {(x + 2)}^{2} {\bigg[\dfrac{x - 1}{x + 2} \bigg]}^{ \dfrac{3}{4} } }$

Now, to evaluate this integral, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:\dfrac{x - 1}{x + 2} = y}$

$\red{\rm :\longmapsto\:\dfrac{x + 2 - 3}{x + 2} = y}$

$\red{\rm :\longmapsto\:1 - \dfrac{3}{x + 2} = y}$

$\red{\rm :\longmapsto\:\dfrac{3}{(x + 2)^{2} }dx = dy}$

So, on substituting these values in above integral, we have

$\rm \: = \: \dfrac{1}{3} \displaystyle\int\rm \frac{dy}{{\bigg(y\bigg) }^{\dfrac{3}{4} }}$

$\rm \: = \: \dfrac{1}{3} \displaystyle\int\rm {\bigg(y\bigg) }^{\dfrac{ - 3}{4} } \: dy$

$\rm \: = \: \dfrac{1}{3}\dfrac{{\bigg(y\bigg) }^{ - \dfrac{3}{4} + 1}}{ - \dfrac{3}{4} + 1} + c$

$\rm \: = \: \dfrac{1}{3}\dfrac{{\bigg(y\bigg) }^{\dfrac{ - 3 + 4}{4}}}{\dfrac{ - 3 + 4}{4}} + c$

$\rm \: = \: \dfrac{1}{3}\dfrac{{\bigg(y\bigg) }^{\dfrac{1}{4}}}{\dfrac{1}{4}} + c$

$\rm \: = \: \dfrac{4}{3}{\bigg( \dfrac{x - 1}{x + 2} \bigg) }^{\dfrac{1}{4} } + c$

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Formula Used

$\boxed{\tt{ \dfrac{d}{dx} \frac{1}{ {x}^{n}} = \frac{ - n}{ {x}^{n + 1} } \: }}$

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

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Learn 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

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{{\bigg(x - 1 \bigg) }^{\dfrac{3}{4}}{\bigg(x + 2\bigg) }^{\dfrac{5}{4} }} \\ \\ can \: \: \: be \: \: \: rewritten \: \: \: as \\ \\ \begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{dx}{{\bigg(x - 1 \bigg) }^{\dfrac{3}{4}}{\bigg(x + 2\bigg) }^{\dfrac{5}{4}} \times {\bigg(x + 2 \bigg) }^{\dfrac{3}{4}} \times \dfrac{1}{{\bigg(x + 2\bigg) }^{\dfrac{3}{4} }} } \\ \end{gathered}$

can be further rewritten as

$\begin{gathered} \rm \: = \: \dfrac{dx}{ {(x + 2)}^{2} {\bigg[\dfrac{x - 1}{x + 2} \bigg]}^{ \dfrac{3}{4} } } \\ \end{gathered}$

Now, to evaluate this integral, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:\dfrac{x - 1}{x + 2} = y} \\ \\ \red{\rm :\longmapsto\:\dfrac{x + 2 - 3}{x + 2} = y} \\ \\ \red{\rm :\longmapsto\:1 - \dfrac{3}{x + 2} = y} \\ \\ \red{\rm :\longmapsto\:\dfrac{3}{(x + 2)^{2} }dx = dy}$

So, on substituting these values in above integral, we have

$\rm \: = \: \dfrac{1}{3} \displaystyle\int\rm \frac{dy}{{\bigg(y\bigg) }^{\dfrac{3}{4} }}</p><p> \\ \\ \begin{gathered} \rm \: = \: \dfrac{1}{3}\dfrac{{\bigg(y\bigg) }^{ - \dfrac{3}{4} + 1}}{ - \dfrac{3}{4} + 1} + c \\ \end{gathered} \\ \\ \begin{gathered} \rm \: = \: \dfrac{1}{3}\dfrac{{\bigg(y\bigg) }^{\dfrac{ - 3 + 4}{4}}}{\dfrac{ - 3 + 4}{4}} + c \\ \end{gathered} \\ \\ \begin{gathered} \rm \: = \: \dfrac{1}{3}\dfrac{{\bigg(y\bigg) }^{\dfrac{1}{4}}}{\dfrac{1}{4}} + c \\ \end{gathered} \\ \\ \begin{gathered} \rm \: = \: \dfrac{4}{3}{\bigg( \dfrac{x - 1}{x + 2} \bigg) }^{\dfrac{1}{4} } + c \\ \end{gathered}$

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Formula Used

$\begin{gathered}\boxed{\tt{ \dfrac{d}{dx} \frac{1}{ {x}^{n}} = \frac{ - n}{ {x}^{n + 1} } \: }} \\ \end{gathered} \\ \\ \\ \begin{gathered}\boxed{\tt{ \displaystyle\int\rm {x}^{n}dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }} \\ \end{gathered}$

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