Question

find integration of :-
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only for class 12 maths students
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Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int {x}^{2} \bigg(1 - \dfrac{1}{ {x}^{2} } + {x}^{5} \bigg) \: dx$

Can be written as:

$\displaystyle \rm \longrightarrow I = \int \bigg( {x}^{2} -1+ {x}^{7} \bigg) \: dx$

$\displaystyle \rm \longrightarrow I = \int{x}^{2}\:dx- \int1 \: dx+ \int {x}^{7} \: dx$

We know that:

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

Using this result, we get:

$\displaystyle \rm \longrightarrow I = \dfrac{ {x}^{2 + 1} }{2 + 1} - \dfrac{ {x}^{0 + 1} }{0 + 1} + \dfrac{ {x}^{7 + 1} }{7 + 1} + C$

$\displaystyle \rm \longrightarrow I = \dfrac{ {x}^{3} }{3} - x+ \dfrac{ {x}^{8} }{8} + C$

$\displaystyle \rm \longrightarrow I = \dfrac{ {x}^{3} }{3}+ \dfrac{ {x}^{8} }{8} - x + C$

Therefore:

$\displaystyle \rm \longrightarrow \int {x}^{2} \bigg(1 - \dfrac{1}{ {x}^{2} } + {x}^{5} \bigg) \: dx = \dfrac{ {x}^{3} }{3}+ \dfrac{ {x}^{8} }{8} - x + C$

★ Which is our required answer.

Learn More:

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