Question

Solve the following


`\sf{\int \: \frac{ {x}^{ (1)/(2) } }{1 + {x}^{ (3)/(4) } } }`
​

Answer 1

Explanation:

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

Given integral is

`\displaystyle\int\rm \frac{{\bigg( x\bigg) }^{(1)/(2) }}{1 + {\bigg( x \bigg) }^{(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) }^{(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) } * {4y}^(3) \:`

`\rm \: \longmapsto= \: 4\displaystyle\int\rm \frac{ {y}^(3) * {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`

`(dt)/(3)`

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

`\rm \: = \: 4\displaystyle\int\rm ((t - 1))/(t) \: * (1)/(3) \:`

`\rm \: = \: (4)/(3) \displaystyle\int\rm ((t - 1))/(t) \:`

`\rm \: = \: (4)/(3) \displaystyle\int\rm \bigg[1 - (1)/(t) \bigg] \:`

`\rm \: = \: (4)/(3) \bigg(t \: - \: log |t|\bigg) + c=34(`

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

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

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ADDITIONAL INFORMATION

`\sf{ \boxed{\begin{array}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 (1)/(x) & \sf logx+ c\\ \\ \sf {e}^(x) & \sf {e}^(x) + c\end{array}} }`