Teknik Pengintegralan (Bagian 1)

1. Integral Substitusi

Ada beberapa bentuk integral yang terkadang pengintegralannya membutuhkan teknik tertentu. Di antara bentuk tertentu itu adalah dengan substitusi, yaitu:

${\displaystyle \int u^n.u^{\prime }dx={\displaystyle \int u^{n}du={\displaystyle \frac{1}{n+1}{u}^{n+1}+C}}}$ .

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {(x^2+3)}^{20}2xdx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& x^2& +& 3& ,& \text{maka}\\ du& =& 2x& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& \int {(x^2+3)}^{20}2xdx=\int u^{20}du\\ & =\frac{1}{21}{u}^{21}+C=\frac{1}{21}{(x^2+3)}^{21}+C\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {(x^4-x^2)}^5(16x^3-8x)dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& x^4& -& x^2& ,& \text{maka}\\ du& =& 4x^3& -& 2x& dx\\ 4du& =& 16x^3& -& 8x& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int u^5.4du=\frac{4}{6}{u}^6+C}\\ & ={\displaystyle \frac{2}{3}{(x^4-x^2)}^6+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int \frac{x+2}{x^2+4x+4}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccccc}u& =& x^2& +& 4x& +& 4& ,& \text{maka}\\ du& =& 2x& +& 4& dx\\ \frac{1}{2}du& =& x& +& 2& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int \frac{1}{u}.\frac{1}{2}du={\displaystyle \frac{1}{2}\int \frac{1}{u}du}}\\ & ={\displaystyle \frac{1}{2}\ln u+C=\frac{1}{2}\ln (x^2+4x+4)+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int \frac{e^{3y}}{{(1-2e^{3y})}^2}dy}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& 1& -& 2e^{3y}& ,& \text{maka}\\ du& =& -& 6e^{3y}& dy& ,\\ -\frac{1}{6}du& =& e^{3y}& dy\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int -\frac{1}{6}\frac{1}{u^2}du=-\frac{1}{6}\int \frac{du}{u^2}}\\ & =-{\displaystyle \frac{1}{6}(-1)\left(u^{-1}\right)+C}\\ & ={\displaystyle \frac{1}{6u}+C}\\ & ={\displaystyle \frac{1}{6}.\frac{1}{1-2e^{3y}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int 12x{(x^2+3)}^{5}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{r}\text{Misalkan}\\ u& =x^2+3\\ du& =2xdx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int 12x{(x^2+3)}^{5}dx}\\ & =\int {(x^2+3)}^5.6.2xdx\\ & =6\int \underset{u^5}{\underbrace{{(x^2+3)}^5}}.\underset{du}{\underbrace{2xdx}}\\ & =6\int u^{5}du\\ & =6.{\displaystyle \frac{u^{5+1}}{5+1}+C}\\ & =u^6+C\\ & ={(x^2+3)}^6+C\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int x\sqrt{x-1}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}u=x-1\\ & du=1dx\iff du=dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int (x-1+1)\sqrt{x-1}dx\\ & =\int (\underset{u}{\underbrace{(x-1)}}+1)\sqrt{\underset{u}{\underbrace{x-1}}}dx\\ & =\int (u+1)\sqrt{u}du\\ & =\int (u\sqrt{u}+\sqrt{u})du\\ & =\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})du\\ & ={\displaystyle \frac{1}{(\frac{3}{2}+1)}{u}^{(\frac{3}{2}+1)}+{\displaystyle \frac{1}{(\frac{1}{2}+1)}{u}^{(\frac{1}{2}+1)}+C}}\\ & ={\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {\displaystyle \frac{x^5}{x^6+a}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x^6+a\\ & \iff du=6x^{5}dx\\ & \iff {\displaystyle \frac{1}{6}du=x^{5}dx}\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& \int {\displaystyle \frac{x^5}{x^6+a}dx}\\ & =\int {\displaystyle \frac{1}{x^6+a}.x^{5}dx}\\ & =\int {\displaystyle \frac{1}{\underset{u}{\underbrace{x^6+a}}}.\underset{\frac{1}{6}du}{\underbrace{x^{5}dx}}}\\ & ={\displaystyle \frac{1}{6}\int \frac{1}{u}du}\\ & ={\displaystyle \frac{1}{6}\ln \left|u\right|+C}\\ & ={\displaystyle \frac{1}{6}\ln |x^6+a|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {\displaystyle (x^2+1){(x^3+3x+2)}^{5}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x^3+3x+2\\ & \iff du=(3x^2+3)dx\\ & \iff du=3(x^2+1)dx\\ & \iff {\displaystyle \frac{1}{3}du=(x^2+1)dx}\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}\int & (x^2+1){(x^3+3x+2)}^{5}dx\\ & =\int {(x^3+3x+2)}^5.(x^2+1)dx\\ & =\int \underset{u^5}{\underbrace{{(x^3+3x+2)}^5}}.\underset{\frac{1}{3}du}{\underbrace{(x^2+1)dx}}\\ & ={\displaystyle \frac{1}{3}\int u^{5}du}\\ & ={\displaystyle \frac{1}{3}.{\displaystyle \frac{u^{5+1}}{5+1}+C}}\\ & ={\displaystyle \frac{1}{18}.u^6+C}\\ & ={\displaystyle \frac{1}{18}{(x^3+3x+2)}^6+C}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{rl}& \text{Selesaikan soal berikut ini}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int {\displaystyle \frac{4x^6+3x^5-8}{x^5}dx}}& \text{f}.& {\displaystyle \int {\displaystyle \frac{{(\sqrt{x}+4)}^3}{\sqrt{x}}dx}}\\ & \text{b}.& {\displaystyle \int {\displaystyle \frac{x^2}{\sqrt{4-x^3}}dx}}& \text{g}.& {\displaystyle \int {\displaystyle \frac{x+3}{\sqrt[5]{{(x^2+6x-1)}^2}}dx}}\\ & \text{c}.& {\displaystyle \int x^3{(x^4+10)}^{{.}^{-\frac{2}{3}}}dx}& \text{h}.& {\displaystyle \int x^5\sqrt{x^6+1}dx}\end{array}\end{array}$.

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.