Question

...........................Evaluate................................​

Answer 1

Question :–

• Calculate the value of :

$\\ \: \:{ \bold{ \int \dfrac{ {e}^{6 log(x) } - {e}^{5 log(x) } }{ {e}^{4 log(x) } - {e}^{3 log(x) } }.dx }}$

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ANSWER :–

$\: \\ \implies{ \bold{I = \int \dfrac{ {e}^{6 log(x) } - {e}^{5 log(x) } }{ {e}^{4 log(x) } - {e}^{3 log(x) } }.dx }}$

• Using identity –

$\: \\ \to \: { \red{ \bold{ log( {x}^{a} ) = a log(x) }}}$

• So that –

$\: \\ \implies{ \bold{I = \int \dfrac{ {e}^{ log( {x}^{6} ) } - {e}^{ log( {x}^{5} ) } }{ {e}^{ log( {x}^{4} ) } - {e}^{ log( {x}^{3} ) } }.dx }}$

• Now using identity –

$\: \\ \to \: { \red{ \bold{ {e}^{ log(x) } = x }}}$

• So that –

$\: \\ \implies{ \bold{I = \int \dfrac{ {x}^{6} - {x}^{5} }{ {x}^{4} - {x}^{3} }.dx }}$

$\: \\ \implies{ \bold{I = \int \dfrac{ {x}^{5} \cancel{(x - 1) } }{ {x}^{3} \cancel{(x - 1)}}.dx }}$

$\: \\ \implies{ \bold{I = \int \dfrac{ {x}^{5} }{ {x}^{3} }.dx }}$

$\: \\ \implies{ \bold{I = \int {x}^{2} .dx }}$

• We know that –

$\: \\ \to { \red{ \bold{ \int {x}^{n } .dx = \dfrac{ {x}^{n + 1} }{n + 1} + c }}}$

• So that –

$\: \\ \implies{ \bold{I = \frac{ {x}^{2 + 1} }{2 + 1} + c}}$

$\: \\ \implies \large{ \pink{ \boxed{ \bold{I = \frac{ {x}^{3} }{3} + c}}}}$

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Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

${\star{\sf{\green{ \int \: \frac{ {e}^{6logx} - {e}^{5logx} }{ {e}^{4logx } - {e}^{3logx} } }}}}$

${\sf{\underline{\purple{Now,}}}}$

${\tt{ I = \int \: \frac{ {e}^{6 \: logx} - {e}^{5 \: logx} }{ {e}^{4 \: logx} - {e}^{3 \: logx} } }}$

$\star{\sf{\underline{\blue{Now,\:we\:know\:that \: log {x}^{a} = a \: logx,}}}}$

${\tt{ I = \int \: \frac{ {e}^{ {logx}^{6} - {e}^{ {logx}^{5} } } }{ {e}^{ {logx}^{4} } - {e}^{ {logx}^{3} } } }}$

${\sf{\underline{\blue{Now, \: {e}^{log(x) } = x }}}}$

${\tt{I = \int \: \frac{ {x}^{6} - {x}^{5} }{ {x}^{4} - {x}^{3} }dx }}$

${\tt{I = \int \: \frac{ {x}^{5}(x - 1) }{ {x}^{3}(x - 1) }dx }}$

${\tt{I = \int \: \frac{ {x}^{5} }{ {x}^{3} }dx }}$

${\tt{I = \int {x}^{5}. {x}^{ - 3} dx }}$

${\tt{I = \int {x}^{5 - 3} dx }}$

${\tt{I = \int {x}^{2} dx }}$

${\tt{I = \bigg [ \frac{ {x}^{2 + 1} }{2 + 1} \bigg ] }} + c$

${\tt{I = \bigg [ \frac{ {x}^{3} }{3} \bigg ] }} + c$

$\boxed{\sf{\underline{\purple{formula \: used.}}}}$

${\tt{ \int \: {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} }} + c \: (\: where \: n \neq \: - 1)</p><p>$