CLA1R0
kak Ram mau nanya boleh ?saya masih penasaran sama kerjaan di KAI namun tidak pada instasinya itu bgmn , dan kata orang2 masuk KAI lulusan SMA itu mitos , sehingga saya dulu pernah mengurungkan niat buat daftar ke KAI :) , kalau ada waktu sempatkan untuk merespon komenan saya ya kak RAM , mohon maaf sebelumnya.
Jawab:
[tex]\displaystyle 7\ln|x+4|-3\ln|x+1|-\ln|2x-1|+C[/tex]
Penjelasan dengan langkah-langkah:
Integration by partial fraction decomposition. Distinct linear factors.
[tex]\displaystyle \int \frac{3(2x^2-8x-1)}{(x+4)(x+1)(2x-1)}~dx=\int \frac{6x^2-24x-3}{(x+4)(x+1)(2x-1)}~dx\\\frac{6x^2-24x-3}{(x+4)(x+1)(2x-1)}=\frac{A}{x+4}+\frac{B}{x+1}+\frac{C}{2x-1}[/tex]
This problem can be solve by Heaviside cover-up method
[tex]\begin{aligned}\frac{6x^2-24x-3}{(x+4)(x+1)(2x-1)}&\:=\frac{A}{x+4}+\frac{B}{x+1}+\frac{C}{2x-1}\\6x^2-24x-3\:&=A(x+1)(2x-1)+B(x+4)(2x-1)+C(x+4)(x+1)\end{aligned}[/tex]
Put x = -4
[tex]\begin{aligned}6x^2-24x-3&\:=A(x+1)(2x-1)+B(x+4)(2x-1)+C(x+4)(x+1)\\6(-4)^2-24(-4)-3\:&=A(-4+1)[2(-4)-1]+B(-4+4)[2(-4)-1]+C(-4+4)(-4+1)\\189\:&=27A\\A\:&=7\end{aligned}[/tex]
Put x = -1
[tex]\begin{aligned}6x^2-24x-3&\:=A(x+1)(2x-1)+B(x+4)(2x-1)+C(x+4)(x+1)\\6(-1)^2-24(-1)-3\:&=A(-1+1)[2(-1)-1]+B(-1+4)[2(-1)-1]+C(-1+4)(-1+1)\\27\:&=-9B\\B\:&=-3\end{aligned}[/tex]
Put x = ½
[tex]\begin{aligned}6x^2-24x-3&\:=A(x+1)(2x-1)+B(x+4)(2x-1)+C(x+4)(x+1)\\6(0.5)^2-24(0.5)-3\:&=A(0.5+1)[2(0.5)-1]+B(0.5+4)[2(0.5)-1]+C(0.5+4)(0.5+1)\\-\frac{27}{2}\:&=\frac{27}{4}~C\\C\:&=-2\end{aligned}[/tex]
So
[tex]\begin{aligned}\int \frac{3(2x^2-8x-1)}{(x+4)(x+1)(2x-1)}~dx&\:=\int \frac{6x^2-24x-3}{(x+4)(x+1)(2x-1)}~dx\\\:&=\int \left ( \frac{7}{x+4}-\frac{3}{x+1}-\frac{2}{2x-1} \right )dx\\\:&=7\ln|x+4|-3\ln|x+1|-\ln|2x-1|+C\end{aligned}[/tex]