Question

Can someone please help me out with this? Please don't spam.
I need step by step explanation.
I will mark the best answer brainliest and thank it.
Please this is a bit urgent.​

Answer 1

GIVEN :–

$\\ \bf \implies \int \dfrac{4 {x}^{3} + \lambda {4}^{x} }{ {4}^{x} + {x}^{4}}dx = log | {4}^{x} + {x}^{4}|$

TO FIND :–

• Value of $\bf \lambda$ = ?

SOLUTION :–

$\\ \bf \implies \int \dfrac{4 {x}^{3} + \lambda {4}^{x} }{ {4}^{x} + {x}^{4}}dx = log | {4}^{x} + {x}^{4}|$

• Now differentiate with respect to 'x' –

$\\ \bf \implies \dfrac{4 {x}^{3} + \lambda {4}^{x} }{ {4}^{x} + {x}^{4}} = \dfrac{d}{dx} \{log | {4}^{x} + {x}^{4}| \}$

• Using identity –

$\\ \bf \to \dfrac{d \{log(x)\}}{dx} = \dfrac{1}{x}$

$\\ \bf \to \dfrac{d( {a}^{x})}{dx} = {a}^{x} ln(x)$

$\\ \bf \to \dfrac{d( {x}^{n})}{dx} = \dfrac{ {x}^{n + 1} }{n + 1}$

• So that –

$\\ \bf \implies \dfrac{4 {x}^{3} + \lambda {4}^{x} }{ {4}^{x} + {x}^{4}} = \dfrac{4 {x}^{3} + {4}^{x} ln(4) }{{4}^{x} + {x}^{4}}$

• Now compare both sides –

$\\ \large\implies { \boxed{ \bf\lambda = ln(4)}}$