Question

Solve the following

Answer 1

Answer

i dont know okay

Answer 2

Step-by-step explanation:

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

Given integral is

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

To evaluate this integral, we have to first remove the fractional exponents from the integrand.

So, we substitute

$\sf{x = {y}^{4} \: \: \: \: \: \: \: \: \: \: \bigg[\rm\implies \:y = {\bigg(x\bigg) }^{\dfrac{1}{4} }\bigg] }$

So, on substituting these values, above integral can be rewritten as

$\rm \: = \longmapsto\: \displaystyle\int\rm \frac{ {y}^{2} }{1 + {y}^{3} } \times {4y}^{3} \:$

$\rm \: \longmapsto= \: 4\displaystyle\int\rm \frac{ {y}^{3} \times {y}^{2} }{1 + {y}^{3} } \:$

To evaluate this integral further, we substitute

$\rm \: \longmapsto \: 1 + {y}^{3} = t1+y3=t$

$\rm \:\longmapsto {y}^{3} = t - 1y3=t−1$

$\sf \: \longmapsto{3y}^{2} \: dy \: = \: dt3y2$

$\dfrac{dt}{3}$

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

$\rm \: = \: 4\displaystyle\int\rm \frac{(t - 1)}{t} \: \times \dfrac{1}{3} \:$

$\rm \: = \: \dfrac{4}{3} \displaystyle\int\rm \frac{(t - 1)}{t} \:$

$\rm \: = \: \dfrac{4}{3} \displaystyle\int\rm \bigg[1 - \frac{1}{t} \bigg] \:$

$\rm \: = \: \dfrac{4}{3} \bigg(t \: - \: log |t|\bigg) + c=34($

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

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

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ADDITIONAL INFORMATION

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