Question

Solve the following

$\int \: \frac{ {x}^{ \frac{1}{2} } }{1 + {x}^{ \frac{3}{4} } } \: dx$
​

Answer 1

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

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

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

$\rm\implies \:dx = {4y}^{3} \: dy$

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

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

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

To evaluate this integral further, we substitute

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

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

$\rm \: {3y}^{2} \: dy \: = \: dt$

$\rm\implies \: {y}^{2} \: dy \: = \: \dfrac{dt}{3}$

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

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

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

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

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

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

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

And Now,

$\bold{ \sf \large \to now} \sf \\ \\ \to \sf \: 4 \int \: \frac{(t - 1)}{t} \times \frac{1}{3} dt \\ \\ \to \sf \: \frac{4}{3} \int \: \frac{(t - 1)}{t} dt \\ \\ \sf \to \: \frac{4}{3} \int \: (1 - \frac{1}{t} )dt \\ \\ \to \sf \: \frac{4}{3} \int \: (t - log |t|)tc \\ \\ \sf \to \: \frac{4}{3} ( {y}^{3} + 1 - log | {y}^{3} + 1 | )tc \\ \\ \boxed{ \large{ \rm \bold{ \green{ \frac{4}{3}((x) ^{ \frac{3}{4} } + 1log |(x) ^{ \frac{3}{4} } + 1| ) + c}}}}$