Integral Fungsi Aljabar

A. Pengertian

Pengintegralan dari suatu fungsi $f(x)$ berbentuk  $\int f(x)dx$ dapat disebut sebagai integral tak tentu dari fungsi  $f(x)$ dan jika  $F(x)$ adalah anti turunan dari  $f(x)$, maka  $F(x)dx=F(x)+C$.

$\begin{array}{rl}& \text{Dengan}\\ & \begin{array}{rl}F(x)& =\text{fungsi integral dari}f(x)\\ f(x)& =\text{fungsi yang diintegralkan}\\ C& =\mathbf{\text{Konstanta}}\end{array}\end{array}$.

B. Rumus Dasar Integral tak Tentu Fungsi Ajabar

Berikut rumus dasar yang perlu diingat

$\begin{array}{rl}\bullet & \int dx=x+C\\ \bullet & \int k.\left({\displaystyle f(x)}\right)dx=k\int f(x)dx\\ \bullet & \int (f(x)\pm g(x))dx=\int f(x)dx+\int g(x)dx\\ \bullet & \int a{x}^{n}dx={\displaystyle \frac{a}{n+1}{x}^{n+1}+C}\end{array}$.

Sebagai rumus-rumus integral yang lain adalah sebagai berikut

$\begin{array}{l}\\ \bullet & {\displaystyle \int a{x}^{n}dx=\frac{a}{n+1}.x^{n+1}+C,\text{dengan}n\ne -1}\\ \bullet & {\displaystyle \int adx=ax+C}\\ \bullet & {\displaystyle \int \frac{1}{x}dx=\int x^{-1}dx=\ln x+C}\\ \bullet & {\displaystyle \int \left|x\right|dx=\frac{1}{2}x\left|x\right|+C}\\ \bullet & {\displaystyle \int \ln xdx=x\ln x-x+C}\\ \bullet & {\displaystyle \int e^{x}dx=e^x+C}\\ \bullet & {\displaystyle \int a^{x}dx=\frac{a^x}{\ln a}+C}\\ \bullet & {\displaystyle \int e^{ax}dx=\frac{1}{a}.e^{ax}+C}\\ \bullet & {\displaystyle \int (x^m+x^n+...+x^p)dx}\\ & ={\displaystyle \int x^{m}dx+\int x^{n}dx+...+\int x^{p}d}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int x^{5}dx}\\ & \text{b}.{\displaystyle \int 2022x^{5}dx}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{1}{x^{2022}}dx}}\\ & \text{d}.{\displaystyle \int 2022ydy}\\ & \text{e}.{\displaystyle \int e^{2022x}dx}\\ & \text{f}.{\displaystyle \int {2022}^{x}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int x^{5}dx=\frac{1}{5+1}.x^{5+1}+C=\frac{1}{6}.x^6+C}\\ & \text{b}.{\displaystyle \int 2022x^{5}dx=\frac{2022}{5+1}.x^{5+1}+C=337x^6+C}\\ & \text{c}.{\displaystyle \int \frac{1}{x^{2022}}dx=\int x^{-2022}dx}\\ & ={\displaystyle \frac{1}{-2022+1}.x^{-2022+1}+C=-\frac{1}{2021}.x^{-2021}+C}\\ & =-{\displaystyle \frac{1}{2021x^{2021}}+C}\\ & \text{d}.{\displaystyle \int 2022ydy=\frac{2022}{1+1}{y}^{1+1}+C=1011y^2+C}\\ & \text{e}.{\displaystyle \int e^{2022x}dx=\frac{1}{2022}.e^{2022x}+C}\\ & \text{f}.{\displaystyle \int {2022}^{x}dx=\frac{{2022}^x}{\ln 2022}+C}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int (3x^2-x+2-\frac{1}{x}+\frac{3}{x^2})dx}\\ & \text{b}.{\displaystyle \int (x^2-2xy+y^2)dx}\\ & \text{c}.{\displaystyle \int |x-1|+|x-2|dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.{\displaystyle }& \int (3x^2-x+2-\frac{1}{x}+\frac{3}{x^2})dx\\ & ={\displaystyle \int 3x^{2}dx-\int xdx+2\int dx-\int \frac{1}{x}dx+3\int \frac{1}{x^2}dx}\\ & ={\displaystyle \frac{3}{2+1}{x}^{2+1}-\frac{1}{1+1}{x}^{1+1}+2x-\ln x+3\left(\frac{1}{-2+1}{x}^{-2+1}\right)+C}\\ & ={\displaystyle \frac{2}{3}{x}^3-\frac{1}{2}{x}^2+2x-\ln x-\frac{3}{x}+C}\end{array}\\ & \begin{array}{rl}\text{b}.{\displaystyle }& \int (x^2-2xy+y^2)dx\\ & ={\displaystyle \int x^{2}dx-2y\int xdx+y^2\int dx}\\ & ={\displaystyle \frac{1}{2+1}{x}^{2+1}-\frac{2y}{1+1}{x}^{1+1}+y^2.x+C}\\ & ={\displaystyle \frac{1}{3}{x}^3-x^{2}y+x{y}^2+C}\end{array}\\ & \begin{array}{rl}\text{c}.{\displaystyle }& \int |x-1|+|x-2|dx\\ & ={\displaystyle \frac{(x-1)}{2}|x-1|+\frac{(x-2)}{2}|x-2|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 2x^{{.}^{\frac{2}{3}}}dx}\\ & \text{b}.{\displaystyle \int \frac{1}{3}\sqrt[4]{x^3}dx}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{x^4-4x^3}{x^2}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int 2x^{{.}^{\frac{2}{3}}}dx={\displaystyle \frac{2}{{\displaystyle \frac{2}{3}+1}}{x}^{{.}^{\frac{2}{3}+1}}+C={\displaystyle \frac{6}{5}{x}^{{.}^{\frac{5}{3}}}+C}}}\\ & \text{b}.{\displaystyle \int \frac{1}{3}\sqrt[4]{x^3}dx={\displaystyle \frac{1}{3}{x}^{{.}^{\frac{3}{4}}}dx=\left({\displaystyle \frac{1}{3}}\right).{\displaystyle \frac{1}{{\displaystyle \frac{3}{4}+1}}{x}^{{.}^{\frac{3}{4}+1}}+C}}}\\ & =\left({\displaystyle \frac{1}{3}}\right){\displaystyle \frac{4}{7}{x}^{{.}^{\frac{7}{4}}}+C={\displaystyle \frac{4}{21}{x}^{{.}^{\frac{7}{4}}}+C}}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{x^4-4x^3}{x^2}dx=\int x^2-4xdx}}\\ & ={\displaystyle \frac{1}{2+1}{x}^{2+1}-{\displaystyle \frac{4}{1+1}{x}^{1+1}+C}}\\ & ={\displaystyle \frac{1}{3}{x}^3-2x^2+C}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{\sqrt{x}(6x^2-2x)}{x}dx}}\\ & \text{b}.{\displaystyle \int (12x^2-4x)(2x+1)dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{\sqrt{x}(6x^2-2x)}{x}dx={\displaystyle \int {\displaystyle \frac{x^{{.}^{\frac{1}{2}}}(6x^2-2x)}{x}dx}}}}\\ & ={\displaystyle \int {\displaystyle \frac{6x^{{.}^{\frac{5}{2}}}-2x^{{.}^{\frac{3}{2}}}}{x}dx={\displaystyle \int 6x^{{.}^{\frac{3}{2}}}-2x^{{.}^{\frac{1}{2}}}dx}}}\\ & ={\displaystyle \frac{6}{{\displaystyle \frac{3}{2}+1}}{x}^{{.}^{\frac{3}{2}+1}}-\frac{2}{{\displaystyle \frac{1}{2}+1}}{x}^{{.}^{\frac{1}{2}+1}}+C}\\ & ={\displaystyle \frac{12}{5}{x}^{{.}^{\frac{5}{2}}}-{\displaystyle \frac{4}{3}{x}^{{.}^{\frac{3}{2}}}+C={\displaystyle \frac{12}{5}{x}^2\sqrt{x}-{\displaystyle \frac{4}{3}x\sqrt{x}+C}}}}\\ & \text{b}.{\displaystyle \int (12x^2-4x)(2x+1)dx}\\ & ={\displaystyle \int (24x^3+4x^2-4x)dx}\\ & ={\displaystyle \frac{24}{3+1}{x}^{3+1}+\frac{4}{2+1}{x}^{2+1}-\frac{4}{1+1}{x}^{1+1}dx}\\ & =6x^4+{\displaystyle \frac{4}{3}{x}^3-2x^2+C}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 2022dx}\\ & \text{b}.{\displaystyle \int \frac{dx}{2022}}\\ & \text{c}.{\displaystyle \int 2022xdx}\\ & \text{d}.{\displaystyle \int -2022x^{2}dx}\\ & \text{e}.{\displaystyle \int (x+2022)dx}\\ & \text{f}.{\displaystyle \int (-2022x^3+2023x^2-2024)dx}\\ & \text{g}.{\displaystyle \int x\sqrt{x}dx}\\ & \text{h}.{\displaystyle \int \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x}}}}}dx}\\ & \text{i}.{\displaystyle \int \frac{2022}{\sqrt[3]{x^2}}dx}\\ & \text{j}.{\displaystyle \int \frac{2022x}{\sqrt[3]{x^5}}dx}\\ & \text{k}.{\displaystyle \int (2022-2020t+t^2)dt}\\ & \text{l}.{\displaystyle \int (\frac{3}{t^3}+\frac{2}{t^2}+2022)dt}\\ & \text{m}.{\displaystyle \int (\sqrt{t}+\frac{1}{2\sqrt{t}})dt}\\ & \text{n}.{\displaystyle \int (a{y}^4+b{y}^2)dy}\\ & \text{o}.{\displaystyle \int (4a{x}^3+3b{x}^2+2cx+1)dx}\\ & \text{p}.{\displaystyle \int \frac{x^2+2022}{x^2}dx}\\ & \text{q}.{\displaystyle \int (e^x+e^{-x})dx}\\ & \text{r}.{\displaystyle \int e^{2023x}dx}\\ & \text{s}.{\displaystyle \int \frac{dx}{e^{2023x}}dx}\\ & \text{t}.{\displaystyle \int \left(\sqrt{{10}^x}\right)dx}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int (2022x-2202)dx}\\ & \text{b}.{\displaystyle \int (x^2-2x-8)dx}\\ & \text{c}.{\displaystyle \int \sqrt{2x}dx}\\ & \text{d}.{\displaystyle \int x^2(x+2)(x-1)dx}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 3x(2x-{\displaystyle \frac{1}{x}})dx}\\ & \text{b}.{\displaystyle \int {\displaystyle \frac{(x^2-4)}{x^2}dx}}\\ & \text{c}.{\displaystyle \int {(2x-{\displaystyle \frac{1}{x^2}})}^{2}dx}\\ & \text{d}.{\displaystyle \int x^2\left({\displaystyle \frac{(x+4)(x-3)}{x}}\right)dx}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{2x^3-3x}{\sqrt{x}}dx}}\\ & \text{b}.{\displaystyle \int {\displaystyle \frac{2}{\sqrt{x}}{(1-\sqrt{x^2})}^{2}dx}}\\ & \text{c}.{\displaystyle \int x^2{(\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}})}^{2}dx}\\ & \text{d}.{\displaystyle \int {\displaystyle \frac{x^2{(2+\sqrt[3]{x^4})}^2}{\sqrt{x}}dx}}\end{array}$.

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.