Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
Given integral is
[tex] \displaystyle\int\rm \frac{{\bigg( x\bigg) }^{\dfrac{1}{2} }}{1 + {\bigg( x \bigg) }^{\dfrac{3}{4} }} \: dx [/tex]
To evaluate this integral, we have to first remove the fractional exponents from the integrand.
So, we substitute
[tex] \sf{x = {y}^{4} \: \: \: \: \: \: \: \: \: \: \bigg[\rm\implies \:y = {\bigg(x\bigg) }^{\dfrac{1}{4} }\bigg] }[/tex]
So, on substituting these values, above integral can be rewritten as
[tex]\rm \: = \longmapsto\: \displaystyle\int\rm \frac{ {y}^{2} }{1 + {y}^{3} } \times {4y}^{3} \: [/tex]
[tex]\rm \: \longmapsto= \: 4\displaystyle\int\rm \frac{ {y}^{3} \times {y}^{2} }{1 + {y}^{3} } \:[/tex]
To evaluate this integral further, we substitute
[tex]\rm \: \longmapsto \: 1 + {y}^{3} = t1+y3=t[/tex]
[tex]\rm \:\longmapsto {y}^{3} = t - 1y3=t−1[/tex]
[tex]\sf \: \longmapsto{3y}^{2} \: dy \: = \: dt3y2[/tex]
[tex]\dfrac{dt}{3}[/tex]
So, on substituting these values in above integral, we get
[tex]\rm \: = \: 4\displaystyle\int\rm \frac{(t - 1)}{t} \: \times \dfrac{1}{3} \: [/tex]
[tex]\rm \: = \: \dfrac{4}{3} \displaystyle\int\rm \frac{(t - 1)}{t} \: [/tex]
[tex]\rm \: = \: \dfrac{4}{3} \displaystyle\int\rm \bigg[1 - \frac{1}{t} \bigg] \: [/tex]
[tex]\rm \: = \: \dfrac{4}{3} \bigg(t \: - \: log |t|\bigg) + c=34([/tex]
[tex]\rm \: = \: \dfrac{4}{3} \bigg( {y}^{3} + 1 \: - \: log | {y}^{3} + 1|\bigg) [/tex]
[tex]\rm \: = \: \dfrac{4}{3} \bigg[ {\bigg(x \bigg) }^{\dfrac{3}{4} } + 1 \: - \: log \bigg| {\bigg(x\bigg) }^{\dfrac{3}{4} } + 1\bigg|\bigg] \: [/tex]
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ADDITIONAL INFORMATION
[tex] \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}} }[/tex]
