Question

Cum se rezolva aceasta integrala?

Answer 1

[tex]\displaystyle\\ \frac{1}{16x^2-9}=\frac{1}{(4x-3)(4x+3)},~\textnormal{vom cauta mai departe sa scriem fractia}\\ \frac{1}{(4x+3)(4x-3)}~\textnormal{ca}~\frac{A}{4x-3}-\frac{B}{4x+3}},~A,B\in\mathbb{R}.\\ \frac{1}{(4x+3)(4x-3)}=\frac{A}{4x-3}-\frac{B}{4x+3} \Longleftrightarrow 1=(4A+4B)x+3A-3B,\\ \textnormal{dar,}~1=0\cdot x+1 \Longleftrightarrow \begin{cases} 4A+4B=0 && 3A-3B=1 \end{cases},\\ \textnormal{de unde obtinem}~A=\frac{1}{6},~B=-\frac{1}{6}.\\[/tex]

[tex]\displaystyle\\ \textnormal{Deci, } \boxed{\frac{1}{16x^2-9}=\frac{1}{6(4x-3)}-\frac{1}{6(4x+3)}}.\\ \textnormal{Revenind la integrala, obtinem ca :}\\ \int \frac{dx}{16x^2-9}=\int \bigg(\frac{1}{6(4x-3)}-\frac{1}{(4x+3)}\bigg) ~dx=\int \frac{1}{6(4x-3)}dx-\int \frac{1}{6(4x+3)}dx.\\ \int \frac{1}{6(4x\pm3)}dx=\frac{1}{6} \int \frac{1}{4x\pm 3}dx\stackrel{u=4x\pm3}= \frac{1}{6}\cdot\frac{1}{4}\int \frac{1}{u} du=\frac{1}{24}\ln |u|.\\ [/tex]

[tex]\displaystyle\\ \textnormal{Asadar,} \boxed{\int \frac{dx}{16x^2-9}=\frac{1}{24}\left( \ln |4x-3|-\ln|4x+3|)+\mathcal{C}}[/tex]