Metode Pecahan Parsial Sebagai Alternatif Metode Ostrogradsky

Ada satu metode yang tidak saya dapati di mata kuliah Matematika Dasar saat memelajari kalkulus integral, yaitu metode Ostrogradsky. Tahu metode ini dari buku KALKULUS penulis H.M. Hasyim Baisuni. Kebetulan ada soal dari buku ini yang melibatkan metode tersebut pada halaman 194 nomor 24. Walau sudah saya pelajari secara otodidak metodenya, saya masih belum paham. Maka dari itu saya selesaikan dengan metode partial fraction decomposition. Soal tersebut ialah ∫ dx / ((x + 1)(x^2 + x + 1)²). Ternyata panjang dan memerlukan ketelitian untuk soal ini.

\displaystyle \frac{1}{(x+1)(x^2+x+1)^2}=\frac{A}{x+1}+\frac{Bx+C}{x^2+x+1}+\frac{Dx+E}{(x^2+x+1)^2}\\ \frac{1}{(x+1)(x^2+x+1)^2}=\frac{A(x^2+x+1)^2+(Bx+C)(x+1)(x^2+x+1)+(Dx+E)(x+1)}{(x+1)(x^2+x+1)^2}\\ 1=Ax^4+2Ax^3+3Ax^2+2Ax+A+Bx^4+2Bx^3+2Bx^2+Bx+Cx^3+2Cx^2+2Cx+C+Dx^2+Dx+Ex+E\\ 1=(A+B)x^4+(2A+2B+C)x^3+(3A+2B+2C+D)x^2+(2A+B+2C+D+E)x+(A+C+E)

A + B = 0 → A = –B

♦ 2A + 2B + C = 0

2(-B) + 2B + C = 0 → C = 0

♦ 3A + 2B + 2C + D = 0

3(-B) + 2B + 2(0) + D = 0

B + D = 0 → B = D

♦ 2A + B + 2C + D + E  =0

2(-B) + B + 2(0) + B + E = 0 → E = 0

A + C + E = 1

A + 0 + 0 = 1 → A = 1B = -1D = -1

\displaystyle \int \frac{dx}{(x+1)(x^2+x+1)^2}=\int \left (\frac{1}{x+1}-\frac{x}{x^2+x+1}-\frac{x}{(x^2+x+1)^2} \right )dx

Selesaikan soal ini menjadi 3 bagian:

  • Bagian pertama: ∫ dx / (x + 1)

Sangat mudah diselesaikan! Dengan metode substitusi diperoleh:

\displaystyle \int \frac{dx}{x+1}\: dx=\ln |x+1|

Ingat! Penggunaan konstanta (C) di penyelesaian akhir.

  • Bagian kedua: ∫ dx / (x² + x + 1)

\displaystyle \int \frac{x}{x^2+x+1}\\ =\int \frac{\frac{1}{2}(2x+1)-\frac{1}{2}}{x^2+x+1}\: dx=\frac{1}{2}\int \left ( \frac{2x+1}{x^2+x+1}-\frac{1}{x^2+x+1} \right )dx

◊ Sub bagian kedua i: ∫ (2x + 1) / (x² + x + 1) dx

\displaystyle \int \frac{2x+1}{x^2+x+1}\\ u=x^2+x+1\rightarrow \frac{du}{dx}=2x+1\\ =\int \frac{du}{u}\\ =\ln |u|=\ln (x^2+x+1)

◊ Sub bagian kedua ii: ∫ dx / (x² + x + 1)

\displaystyle \int \frac{dx}{x^2+x+1}\\ =\int \frac{dx}{x^2+x+\left ( \frac{1}{2} \right )^2-\left ( \frac{1}{2} \right )^2+1}\\ =\int \frac{dx}{\left ( x+\frac{1}{2} \right )^2+\frac{3}{4}}

Berdasarkan rumus \displaystyle \int \frac{du}{a^2+u^2}=\frac{\tan^{-1}\left ( \frac{u}{a} \right )}{a}+C, a>0, dan misal a² = ¾ & u = x + ½, maka:

\displaystyle =\frac{\tan^{-1}\left ( \frac{x+\frac{1}{2}}{\sqrt{\frac{3}{4}}} \right )}{\sqrt{\frac{3}{4}}}=\frac{2\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{\sqrt{3}}

Jadi:

\displaystyle \int \frac{x}{x^2+x+1}dx\\ =\frac{1}{2}\int \left ( \frac{2x+1}{x^2+x+1}-\frac{1}{x^2+x+1} \right )dx\\ =\frac{1}{2}\left [ \ln (x^2+x+1)-\frac{2\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{\sqrt{3}} \right ]\\ =\frac{\ln (x^2+x+1)}{2}-\frac{\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{\sqrt{3}}

  • Bagian ketiga: ∫ dx / (x² + x + 1)²

Sama seperti pada bagian kedua: \displaystyle \int \frac{x}{(x^2+x+1)^2}dx=\frac{1}{2}\int \left [\frac{2x+1}{(x^2+x+1)^2}-\frac{1}{(x^2+x+1)^2} \right ]dx

◊ Sub bagian ketiga i: ∫ (2x + 1) / (x² + x + 1)² dx

Dengan metode substitusi diperoleh:

\displaystyle \int \frac{2x+1}{(x^2+x+1)^2}\: dx=-\frac{1}{x^2+x+1}

◊ Sub bagian ketiga ii: ∫ dx / (x² + x + 1)²

\displaystyle \int \frac{dx}{(x^2+x+1)^2}\: dx\\ =\int \frac{dx}{\left [\left ( x+\frac{1}{2} \right )^2+\frac{3}{4} \right ]^2}\\ =\int \left [\frac{1}{\left (x+\frac{1}{2} \right )^2+\frac{3}{4}} \right ]^2dx\\ =\int \left [\frac{4}{4\left (x^2+x+\frac{1}{4}+\frac{3}{4} \right )} \right ]^2dx\\ =16\int \frac{dx}{[(2x+1)^2+3]^2}\\ u=2x+1\rightarrow \frac{du}{dx}=2\\ =8\int \frac{du}{(u^2+3)^2}

Ingat kembali rumus reduksi \displaystyle \int \frac{dx}{(ax^2+b)^n}=\frac{2n-3}{2b(n-1)}\int \frac{dx}{(ax^2+b)^{n-1}}+\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+C, sehingga diperoleh:

\displaystyle =8\left [ \frac{u}{6(u^2+3)}+\frac{1}{6}\int \frac{du}{u^2+3} \right ]\\ =\frac{4u}{3(u^2+3)}+\frac{4}{3}\frac{\tan^{-1}\left ( \frac{u}{3} \right )}{\sqrt{3}}\\ =\frac{4(2x+1)}{3[(2x+1)^2+3]}+\frac{4\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{3\sqrt{3}}

Jadi:

\displaystyle \int \frac{x}{(x^2+x+1)^2}dx\\ =\frac{1}{2}\int \left [\frac{2x-1}{(x^2+x+1)^2}-\frac{1}{(x^2+x+1)^2} \right ]dx\\ =\frac{1}{2}\left \{ -\frac{1}{x^2+x+1}-\left [ \frac{4(2x+1)}{3[(2x+1)^2+3]}+\frac{4\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{3\sqrt{3}} \right ] \right \}\\ =-\frac{1}{2(x^2+x+1)}-\frac{2(2x+1)}{3[(2x+1)^2+3]}-\frac{2\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{3\sqrt{3}}

Jawaban dari soal ini adalah:

\displaystyle \int \frac{dx}{(x+1)(x^2+x+1)^2}\\ =\int \left [\frac{1}{x+1}-\frac{x}{x^2+x+1}-\frac{x}{(x^2+x+1)^2} \right ]dx\\ =\ln |x+1|-\left [ \frac{\ln (x^2+x+1)}{2}-\frac{\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{\sqrt{3}} \right ]-\left [-\frac{1}{2(x^2+x+1)}-\frac{2(2x+1)}{3[(2x+1)^2+3]}-\frac{2\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{3\sqrt{3}} \right ]+C\\ =\ln |x+1|-\frac{\ln (x^2+x+1)}{2}+\frac{5\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{3\sqrt{3}}+\frac{1}{2(x^2+x+1)}+\frac{4x+2}{12(x^2+x+1)}+C\\ =\ln |x+1|-\frac{\ln (x^2+x+1)}{2}+\frac{5\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{3\sqrt{3}}+\frac{6+4x+2}{12(x^2+x+1)}+C\\ =\ln |x+1|-\frac{\ln (x^2+x+1)}{2}+\frac{5\sqrt{3}\tan^{-1}\left ( \frac{2x+1}{\sqrt{3}} \right )}{9}+\frac{x+2}{3(x^2+x+1)}+C

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