Question

plzxz solve this .....................​

Answer 1

Question :-

Integrate it -

$\longmapsto \: \int \sf \frac{ {x}^{9} }{ {(4 {x}^{2} + 1)}^{6} }$

Answer :-

$\to \boxed{ \sf \: \frac{1}{10 {(4 + \frac{1}{ {x}^{2} }) }^{5} } + c \: } \\ \\ or \: \\ \to \boxed{\sf \: \frac{ {x}^{10} }{10 {(4 {x}^{2} + 1) }^{5} } \: + c \: }m$

Solution :-

We have ,

$\to \: \int \sf \frac{ {x}^{9} }{ {(4 {x}^{2} + 1)}^{6} } \\ \: \\ \sf taking \: common \: {x}^{2} \: in \: denomenator \\ \\ \to \: \int \sf \frac{ {x}^{9} }{ { \big \{{x}^{2} (4 + \frac{1}{ {x}^{2} } ) \big \}}^{6} } \\ \: \\ \\ \to \: \int \sf \frac{ {x}^{9} }{ { {x}^{12} (4 + \frac{1}{ {x}^{2} } )}^{6} } \\ \: \\ \to \: \int \sf \frac{ 1 }{ { {x}^{3} (4 + \frac{1}{ {x}^{2} } )}^{6} } \: \\ \\ \sf \: let \: \: 4 + \frac{1}{ {x}^{2} } = t \\ \\ \to \sf \: 4 + {x}^{ - 2} = t \\ \\ \sf differentiate \: with \: respect \: to \: x \\ \\ \to \sf 0 - 2 {x}^{ - 2 - 1} = \frac{dx}{dt} \\ \\ \to \sf \frac{ - 2}{ {x}^{3} } \: dx = dt \\ \: \\ \to \sf \frac{1}{ {x}^{3} } dx = \frac{ - 1}{2} dt \\ \\ \therefore \\ \\ \to \sf \frac{ - 1}{2} \: \int \: \frac{1}{ {t}^{6} } \: dt \\ \\ \to \sf \: \frac{ - 1}{2} \: \int {t}^{ - 6} \: dt \\ \\ \because \sf \: \int \: {x}^{n} dx \: = \frac{ {x}^{n + 1} }{n + 1} \: + c \: (constant \: )$

$\therefore \: \\ \to \sf \: \frac{ - 1}{2} \bigg( \frac{ {t}^{ - 6 + 1} }{ - 6 + 1} \bigg) + c \: \\ \\ \to \sf \: \frac{ - 1}{2} \bigg( \frac{ {t}^{ - 5} }{ - 5} \bigg) \: + c \\ \\ \to \sf \: \frac{1}{10 \: {t}^{5} } \: + c \\ \\ \sf \: now \: put \: t \: = 4 + \frac{1}{ {x}^{2} } \: from \: eq.(1) \\ \\ \to \boxed{ \sf \: \frac{1}{10 {(4 + \frac{1}{ {x}^{2} }) }^{5} } + c \: } \\ \\ or \: \\ \to \boxed{\sf \: \frac{ {x}^{10} }{10 {(4 {x}^{2} + 1) }^{5} } \: + c \: }$