Question

find integration of this
​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{ {x}^{2} - {x}^{4} - {x}^{3/2} }{x} \: dx$

Can be written as:

$\displaystyle \rm \longrightarrow I = \int (x - {x}^{3} - {x}^{3/2 - 1} )\: dx$

$\displaystyle \rm \longrightarrow I = \int (x - {x}^{3} - {x}^{1/2} )\: dx$

$\displaystyle \rm \longrightarrow I = \int (x - {x}^{3} - \sqrt{x} )\: dx$

$\displaystyle \rm \longrightarrow I = \int x \: dx- \int {x}^{3} \: dx - \int\sqrt{x} \: dx$

We know that:

$\displaystyle \rm \longrightarrow \int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} + C$

Using this result, we get:

$\displaystyle \rm \longrightarrow I = \dfrac{ {x}^{1 + 1} }{1 + 1} - \dfrac{ {x}^{3 + 1} }{3 + 1}- \dfrac{ {x}^{1/2 + 1} }{1/2 + 1} +C$

$\displaystyle \rm \longrightarrow I = \dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{4} }{4}- \dfrac{ x \sqrt{x}}{3/2} +C$

$\displaystyle \rm \longrightarrow I = \dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{4} }{4}- \dfrac{ 2x \sqrt{x}}{3} +C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{ {x}^{2} - {x}^{4} - {x}^{3/2} }{x} \: dx= \dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{4} }{4}- \dfrac{ 2x \sqrt{x}}{3} +C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$