Question

Integral de ∫1+4x/4x+8x² ?

Answer 1

Preguiça de fazer procurar simbolo de integral, então vou usar um S

Iremos colocar 4x em evidência, no denominador:

S(1 + 4x/4x(1 + 4x))dx

Cancele 1 + 4x:

S(1/4x)dx = ln(módulo de 4x)

Answer 2

$\displaystyle\mathsf{\int\dfrac{1+4x}{4x+8x^2}dx=\int\dfrac{1+4x}{4x(1+2x)}dx}$

$\mathsf{\dfrac{1+4x}{4x(1+2x)}=\dfrac{A}{4x}+\dfrac{B}{1+2x}}\\\mathsf{\dfrac{1+4x}{4x(1+2x)}=\dfrac{A(1+2x)+B.4x}{4x(1+2x)}}\\\mathsf{A+2Ax+4Bx=1+4x}\\\mathsf{A+(2A+4B)x=1+4x}$

$\begin{cases}\mathsf{A=1}\\\mathsf{2A+4B=4}\end{cases}$

$\mathsf{2A+4B=4}\\mathsf{2.1+4B=4}\\\mathsf{4B=4-2}\\\mathsf{4B=2\implies~B=\dfrac{2}{4}=\dfrac{1}{2}}$

$\displaystyle\mathsf{\int\dfrac{1+4x}{4x+8x^2}=\dfrac{1}{4}\int\dfrac{dx}{x}+\dfrac{1}{2}\int\dfrac{dx}{1+2x}}\\=\mathsf{\dfrac{1}{4}\ell n|x|+\dfrac{1}{2}\ell n|1+2x|+k}$