Question

plssas solve=====,


plsssssz​

Answer 1

Given Question :- Evaluate

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

Given integral is

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

can be rewritten as

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

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

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

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

So, Substitute

$\red{\rm \: 1 + \dfrac{1}{ {x}^{4} } = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm \:0 + \dfrac{ - 4}{ {x}^{5} } = \frac{dy}{dx} }$

$\red{\rm \: - \: \dfrac{4}{ {x}^{5} } = \frac{dy}{dx} }$

$\red{\rm \: \: \dfrac{dx}{ {x}^{5} } = - \frac{1}{4}dy}$

So, on substituting these values, in above expression, we get

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

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

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

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

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

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

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

$\rule{190pt}{2pt}$

Additional information :-

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