Teknik Pengintegralan (Bagian 2)

2. Integral Parsial

2. 1 Integral Parsial

Jika teknik pada no.1 pada pembahasan sebelumnya tidak dapat digunakan, maka kemungkinan adalah dengan menggunakan teknik yang satunya ini, yaitu teknik integral parsial. Adapun untuk teknik integral ini diformulasikan dengan bentuk rumus

${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Perhatian kembali soal berikut}\\ & {\displaystyle \int x\sqrt{x-1}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misalkan}u=x-1\\ & du=1dx\iff du=dx\end{array}\\ & \text{Dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int (x-1+1)\sqrt{x-1}dx\\ & =\int (\underset{u}{\underbrace{(x-1)}}+1)\sqrt{\underset{u}{\underbrace{x-1}}}dx\\ & =\int (u+1)\sqrt{u}du\\ & =\int (u\sqrt{u}+\sqrt{u})du\\ & =\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})du\\ & ={\displaystyle \frac{1}{(\frac{3}{2}+1)}{u}^{(\frac{3}{2}+1)}+{\displaystyle \frac{1}{(\frac{1}{2}+1)}{u}^{(\frac{1}{2}+1)}+C}}\\ & ={\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}\\ \\ & \text{Alternatif 2}\\ & \begin{array}{l}\\ {\displaystyle \int x\sqrt{x-1}dx=\int \underset{{\displaystyle \underset{u}{\mid }}}{x}.\underset{{\displaystyle \underset{dv}{\mid }}}{\underbrace{\sqrt{x-1}dx}}}& \\ \begin{array}{cccc}u=x& & & dv=\sqrt{x-1}dx\\ du=dx& & & {\displaystyle \int dv=\int \sqrt{x-1}dx}\\ & & & v=\int {(x-1)}^{\frac{1}{2}}dx\\ & & & v={\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\end{array}\end{array}\\ & \text{Dengan integral parsial}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int \underset{{\displaystyle \underset{u}{\mid }}}{x}.\underset{{\displaystyle \underset{dv}{\mid }}}{\underbrace{\sqrt{x-1}dx}}=u.v-\int v.du\\ & =x.\left({\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\right)-\int \left({\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\right)dx\\ & ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{2}{3}\times \frac{2}{5}{(x-1)}^{\frac{5}{2}}+C}}\\ & ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\end{array}\end{array}\end{array}$.

$\begin{array}{rl}& \text{Perhatikan bahwa}\\ & \begin{array}{|cc|}\hline \text{Hasil dengan Substitusi}& \text{Hasil dengan Integral Parsial}\\ {\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}& {\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\\ \hline\end{array}\\ & \text{Jika kita sejajarkan dengan ruas }\\ & \text{yang berbeda, maka}\\ & \begin{array}{rl}{\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}& ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\\ {\displaystyle \frac{2}{5}(x-1){(x-1)}^{\frac{3}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}}& ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}(x-1){(x-1)}^{\frac{3}{2}}}}\\ \left({\displaystyle \frac{2}{5}(x-1)+\frac{2}{3}}\right){(x-1)}^{\frac{3}{2}}& =\left({\displaystyle \frac{2x}{3}-\frac{4}{15}(x-1)}\right){(x-1)}^{\frac{3}{2}}\\ {\displaystyle \frac{2x}{5}-\frac{2}{5}+\frac{2}{3}}& ={\displaystyle \frac{2x}{3}-\frac{4x}{15}+\frac{4}{15}}\\ {\displaystyle \frac{2x}{5}+\frac{4}{15}}& ={\displaystyle \frac{2x}{5}+\frac{4}{15}}\\ \text{ruas kiri}& =\text{ruas kanan}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah integral dari}\int (x+2)dx\\ & \text{dengan cara}\\ & \text{a})\text{substitusi}\\ & \text{b})\text{parsial}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \mathbf{\text{Cara substitusi}}\\ & \int (x+2)dx=........?\\ & \text{Misalkan}m=x+2\\ & dm=dx\\ & \text{maka},\\ & \int (x+2)dx=\int \underset{\begin{array}{c}|\\ \mathbf{\text{m}}\end{array}}{\underbrace{(x+2)}}.\underset{\begin{array}{c}|\\ \mathbf{\text{dm}}\end{array}}{\underbrace{dx}}\\ & =\frac{1}{2}{m}^2+C=\frac{1}{2}(x+2{)}^2+C\\ & =\frac{1}{2}(x^2+4x+4)+C=\frac{1}{2}{x}^2+2x+\underset{\begin{array}{c}|\\ \mathbf{\text{C}}\end{array}}{\underbrace{2+C}}\\ & =\frac{1}{2}{x}^2+2x+C\end{array}\\ \\ & \begin{array}{rl}& \mathbf{\text{Cara parsial}}\\ & \int (x+2)dx=\int \underset{\begin{array}{c}|\\ \mathbf{\text{u}}\end{array}}{\underbrace{1}}.\underset{\begin{array}{c}|\\ \mathbf{\text{dv}}\end{array}}{\underbrace{(x+2)dx}}\\ & \{\begin{array}{c}u=1\to du=0\\ \\ v=\underset{\begin{array}{c}|\\ =\frac{1}{2}{x}^2+2x+C\end{array}}{\int dv}\leftarrow dv=(x+2)dx\end{array}\\ & =\mathbf{\text{u.v}}-\int \mathbf{\text{v.du}}\\ & =1.(\frac{1}{2}{x}^2+2x+C)-\int (\frac{1}{2}{x}^2+2x+C).0\\ & =\frac{1}{2}{x}^2+2x+C\end{array}\end{array}$.

2. 2 Aturan Tanzalin

Sumber Referensi

$\begin{array}{rl}& \text{Tentukanlah hasil integral berikut}\\ & {\displaystyle \int 2x^2(x-4{)}^{2}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Bentuk}{\displaystyle \int 2x^2(x-4{)}^{2}dx\text{dianggap sebagai}}\\ & {\displaystyle \int udv\text{dengan}u\text{adalah bagian yang mudah}}\\ & \text{kita diferensialkan, maka}u\text{kita pilihkan}\\ & \text{yaitu}u=2x^2\\ & \text{Selanjutnya dengan}\mathbf{\text{aturan Tanzalin}}\\ & \text{sebagai berikut}\end{array}$.

$\begin{array}{|cc|}\hline \text{Diderensialkan}& \text{Diintegralkan}\\ \begin{array}{rl}& +2x^2\\ & -4x\\ & +4\\ & -0\\ & \end{array}& \begin{array}{rl}& (x-4{)}^4\\ & {\displaystyle \frac{1}{5}(x-4{)}^5}\\ & {\displaystyle \frac{1}{30}(x-4{)}^6}\\ & {\displaystyle \frac{1}{210}(x-4{)}^7}\end{array}\\ \hline\end{array}$.

Hasil dari integral teknik ini adalah:

$\begin{array}{rl}& {\displaystyle \int 2x^2(x-4{)}^{4}dx}\\ & =(+2x^2)\left({\displaystyle \frac{1}{5}(x-4{)}^5}\right)+(-4x)\left({\displaystyle \frac{1}{30}{(x-4)}^6}\right)\\ & +(+4)\left({\displaystyle \frac{1}{210}(x-4{)}^7}\right)+C\\ & ={\displaystyle \frac{2}{5}{x}^2(x-4{)}^5-{\displaystyle \frac{2}{15}x(x-4{)}^6+{\displaystyle \frac{2}{105}(x-4{)}^7+C}}}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{rl}& \text{1. Selesaikan soal berikut ini}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int {\displaystyle x^2\sqrt{x+7}dx}}& \text{d}.& {\displaystyle \int -(2x^2+1)\sqrt{3-x}dx}\\ & \text{b}.& {\displaystyle \int 2x^2\sqrt{3-x}dx}& \text{e}.& {\displaystyle \int x^5(2x+1{)}^{6}dx}\\ & \text{c}.& {\displaystyle \int 3x^2\sqrt{2x+1}dx}& \text{f}.& {\displaystyle \int {\displaystyle \frac{2x^6}{\sqrt[3]{3x-1}}dx}}\end{array}\end{array}$ .

$\begin{array}{rl}& \text{2. Selesaikan soal berikut dengan Metode Tanzalin}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int -(2x^2+1)\sqrt{3-x}dx}\\ & \text{b}.& {\displaystyle \int x^5(2x+1{)}^{6}dx}\\ & \text{c}.& {\displaystyle \int {\displaystyle \frac{2x^6}{\sqrt[3]{3x-1}}dx}}\\ & \text{d}.& {\displaystyle \int {\displaystyle 3(x-2{)}^4.\sqrt[3]{x}dx}}\\ & \text{e}.& {\displaystyle \int {\displaystyle \frac{(x^4-3)}{\sqrt{1-x}}dx}}\\ & \text{f}.& {\displaystyle \int {\displaystyle x^3\sqrt{1-2x}dx}}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.