XI 8.3 : Integral Substitusi

Halo halo hai … Sebelumnya kita sudah membahas beberapa sifat integral tak tentu. Jika kamu masih belum paham beberapa hal, silakan akses materinya di link berikut > https://bdr.masbied.com/bdr-genap-kelas-xi/2712/xi-8-2-sifat-sifat-integral/

Kali ini kita akan membahas tentang integral substitusi.

C. Integral Substitusi

$f(g(x))$ jika kita turunkan terhadap $x$ maka akan kita dapatkan $\frac{d}{dx}f(g(x))=f'(g(x)).g'(x)$. Ini yang kemarin kita bahas sebagai TURUNAN LUAR $\times$ TURUNAN DALAM yah …

Nah, jika $f'(g(x)).g'(x)$ kita integralkan $\int f'(g(x)).g'(x)$, maka hasilnya adalah $f(g(x))+C$.

Ingat kembali, bahwa integral atau anti turunan adalah kebalikan dari turunan.

Kita bisa membuat pemisalan.

$\int f'(g(x)).\textcolor{red}{g'(x)}$

Misalkan :

$u=g(x)$

$\frac{du}{dx}=\textcolor{red}{g'(x)}$, hasil turunan ini terdapat dalam integral.

Kedua ruas kita integralkan terhadap $x$

$\begin{aligned}\int \frac{du}{dx}&= \int \textcolor{red}{g'(x)} dx & \to \frac{du}{dx}=u+C \end{aligned}$

$u + C = \int g'(x) dx$

$\int du = \int g'(x) dx$

Jika disubstitusikan ke bentuk awalnya, maka akan kita dapatkan seperti ini

$\boxed{\int f'(u).du = F(u)+C}$

Inilah yang disebut sebagai integral substitusi.

Pusing kan?

Wkwkw …

Supaya gampang, langsung masuk ke contoh saja deh ..

Contoh 1 :

$\int (x-2)^5dx =\cdots$

Jawab :

$\int (\textcolor{blue}{x-2})^5dx =\cdots$

Misalkan : $\boxed{\textcolor{blue}{u=x-2}}$

$\frac{du}{dx}=1 \gets$ turunan dari $x-2=1$

Karena $\frac{du}{dx}=1 \to \begin{equation}\boxed{\textcolor{red}{du=dx}}\end{equation}$, akibatnya..

$\begin{aligned}\int (\textcolor{blue}{x-2})^5\textcolor{red}{dx} &=\int \textcolor{blue}{u}^5\textcolor{red}{du} \\ &= \frac{1}{5+1}u^{5+1}+C \\ &= \frac{1}{6}u^6 +C & \to u=x-2 \\ &=\frac{1}{6}(x-2)^6+C \end{aligned}$

$\therefore \int (x-2)^5dx=\frac{1}{6}(x-2)^6+C$

Contoh 2 :

$\int (2x+1)^5 dx = \cdots$

Jawab :

$\int (\textcolor{blue}{2x+1})^5 dx$

Misalkan : $\begin{equation}\boxed{\textcolor{blue}{u=2x+1}}\end{equation}$

$\frac{du}{dx} = 2 \to$ Ingat, turunan dari $2x+1$ adalah $2$.

$du= 2dx \to \frac{du}{2}=dx$, yang equivalen dengan

$\begin{equation}\boxed{ \textcolor{red}{dx=\frac{1}{2}du}}\end{equation}$, akibatnya..

$\begin{aligned}\int (\textcolor{blue}{2x+1})^5 \textcolor{red}{dx}&=\int \textcolor{blue}{u}^5.\textcolor{red}{\frac{1}{2}du} \\ &= \frac{1}{2} \int u^5.du \\ &= \frac{1}{2}.\frac{1}{5+1}u^{5+1}+C \\ &= \frac{1}{2}.\frac{1}{6}u^{6}+C\\ &=\frac{1}{12}u^6+C &\to u=\textcolor{blue}{2x+1} \\ &=\frac{1}{12}(2x+1)^6+C \end{aligned}$

$\therefore \int (2x+1)^5 dx=\frac{1}{12}(2x+1)^6+C$

Contoh 3 :

$\int (3x-2)^6 dx = \cdots$

Jawab :

$\int (\textcolor{blue}{3x-2})^6 dx$

Misalkan : $\begin{equation}\boxed{\textcolor{blue}{u=3x-2}}\end{equation}$

$\frac{du}{dx} = 3$

$du=3dx \to \frac{du}{3}=dx$, yang equivalen dengan

$\begin{equation}\boxed{\textcolor{red}{dx=\frac{1}{3}du}}\end{equation}$, akibatnya ..

$\begin{aligned}\int (\textcolor{blue}{3x-2})^6 \textcolor{red}{dx}&=\int \textcolor{blue}{u}^6.\textcolor{red}{\frac{1}{3}du} \\ &= \frac{1}{3} \int u^6.du \\ &= \frac{1}{3}.\frac{1}{6+1}u^{6+1}+C \\ &=\frac{1}{3}.\frac{1}{7}u^7+C \\ &=\frac{1}{21}u^7+C &\to \textcolor{blue}{u=3x-2} \\ &=\frac{1}{21}(3x-2)^7+C \end{aligned}$

$\therefore \int (3x-2)^6 dx=\frac{1}{21}(3x-2)^7+C$

Contoh 4 :

$\int (3\sqrt{5x-2}dx=\cdots$

Jawab :

Misalkan : $\begin{equation}\boxed{\textcolor{blue}{u=5x-2}}\end{equation}$

$\frac{du}{dx}=5$

$du = 5dx \to \frac{1}{5}du = dx$, yang equivalen dengan

$\begin{equation}\boxed{\textcolor{red}{dx=\frac{1}{5}du}}\end{equation}$, akibatnya..

$\begin{aligned}\int (3 \sqrt{\textcolor{blue}{5x-2}}) \textcolor{red}{dx}&=\int 3. (\textcolor{blue}{5x-2})^\frac{1}{2} \textcolor{red}{dx} \\ &= \int 3.\textcolor{blue}{u}^\frac{1}{2}.\textcolor{red}{\frac{1}{5}du} \\ &=3.\frac{1}{5}.\int u^\frac{1}{2}du \\&=\frac{3}{5}.\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}+C \\ &= \frac{3}{5}.\frac{1}{\frac{3}{2}}u^\frac{3}{2}+C \\ &=\frac{3}{5}.\frac{2}{3}u^\frac{3}{2}+C \\ &=\frac{6}{15}u^\frac{3}{2}+C & \gets u=5x-2 \\ &=\frac{2}{5}(5x-2)^\frac{3}{2}+C \end{aligned}$

$\therefore \int (3\sqrt{5x-2}dx=\frac{2}{5}(5x-2)^\frac{3}{2}+C$

Contoh 5 :

$\int (2x(x^2-5)^7)dx =\cdots$

Jawab :

Misalkan : $\boxed{\textcolor{blue}{u=x^2-5}}$

$\frac{du}{dx}=2x$, yang equivalen dengan

$\begin{equation}\boxed{\textcolor{red}{du=2xdx}}\end{equation}$, akibatnya..

$\begin{aligned}\int (2x(\textcolor{blue}{x^2-5})^7)dx &= \int (\textcolor{blue}{x^2-5})^7. \textcolor{red}{2xdx} \\ &= \int \textcolor{blue}{u}^7.\textcolor{red}{du} \\ &=\frac{1}{8}u^8 + C &\gets u=x^2-5 \\ &=\frac{1}{8}(x^2-5)^8+C \end{aligned}$

$\therefore \int (2x(x^2-5)^7)dx=\frac{1}{8}(x^2-5)^8+C$

Rumus Umum $\int (ax+b)^n dx$

Berdasarkan hasil pengerjaan pada contoh 2 & 3, kita dapat simpulkan rumus umum untuk $\int (ax+b)^n dx$ sebagai berikut :

$\int (\textcolor{green}{a}x+b)^\textcolor{red}{n}dx=\frac{1}{\textcolor{green}{a}(\textcolor{red}{n}+1)}(\textcolor{green}{a}x+b)^{\textcolor{red}{n}+1}+C$

Rumus ini dapat digunakan dengan syarat :

• Variabel $x$ linear (atau berpangkat $1$)
• Variabel yang ada hanya $x$ saja.
• Pangkat $(n) \neq -1$

Contoh 6 :

$\int 2(3x+8)^5 dx =\cdots$

Jawab :

$\begin{aligned}\int 2(\textcolor{green}{3}x+8)^\textcolor{red}{5} dx &= 2.\int (\textcolor{green}{3}x+8)^\textcolor{red}{5} dx \\ &=2. \frac{1}{\textcolor{green}{3}(\textcolor{red}{5}+1)}(\textcolor{green}{3}x+8)^{\textcolor{red}{5}+1}+C \\ &=2.\frac{1}{3(6)}.(3x+8)^6+C \\ &=2.\frac{1}{18}.(3x+8)^6+C \\ &=\frac{2}{18}.(3x+8)^6+C \\ &=\frac{1}{9}(3x+8)^6+C \end{aligned}$

$\therefore \int 2(3x+8)^5 dx=\frac{1}{9}(3x+8)^6+C$

Contoh 7 :

$\int 5(7x+1)^4 dx =\cdots$

Jawab :

$\begin{aligned}\int 5(\textcolor{green}{7}x+1)^\textcolor{red}{4} dx &= 5.\int (\textcolor{green}{7}x+1)^\textcolor{red}{4} dx \\ &=5. \frac{1}{\textcolor{green}{7}(\textcolor{red}{4}+1)}(\textcolor{green}{7}x+1)^{\textcolor{red}{4}+1}+C \\ &=5.\frac{1}{7(5)}.(7x+1)^5+C \\ &=5.\frac{1}{35}.(7x+1)^5+C \\ &=\frac{5}{35}.(7x+1)^5+C \\ &=\frac{1}{7}(7x+1)^5+C \end{aligned}$

$\therefore \int 5(7x+1)^4 dx=\frac{1}{7}.(7x+1)^5+C$

Contoh 8 :

$\int [(4x+4)(x^2+2x)^7]dx=\cdots$

Jawab :

Misalkan :

$\begin{equation}\boxed{\textcolor{blue}{u=x^2+2x}}\end{equation}$

$\frac{du}{dx}=2x+2$

$du=(2x+2)dx$, yang equivalen dengan

$\begin{equation}\boxed{\textcolor{red}{2du = (4x+4)dx}} \end{equation}\gets$ kedua ruas $\times 2$

Note : Bentuk $(2x+2)$ dimanipulasi menjadi $(4x+4)$, karena bentuk $(4x+4)$ terdapat dalam soal.

Akibatnya..

$\begin{aligned}\int [(4x+4)(x^2+2x)^7]dx &= \int \textcolor{blue}{(x^2+2x)}^7.\textcolor{red}{(4x+4)dx} \\ &= \int \textcolor{blue}{u}^7.\textcolor{red}{2du} \\ &= 2.\int u^7.du \\ &= 2. \frac{1}{8}u^8+C \\ &=\frac{1}{4}u^8+C &\gets u=x^2+2x \\ &= \frac{1}{4}(x^2+2x)^8+C \end{aligned}$

$\therefore \int [(4x+4)(x^2+2x)^7]dx=\frac{1}{4}(x^2+2x)^8+C$

Contoh 9 :

$\int (\frac{2x+1}{\sqrt{x^2+x-6}})dx =\cdots$

Jawab :

Misalkan : $\begin{equation}\boxed{\textcolor{blue}{u=x^2+x-6}}\end{equation}$

$\frac{du}{dx}=2x+1$

$\begin{equation}\boxed{\textcolor{red}{du=(2x+1)dx}} \end{equation} \gets$ kedua ruas diintegralkan

Akibatnya..

$\begin{aligned}\int (\frac{\textcolor{red}{2x+1}}{\sqrt {\textcolor{blue}{x^2+x-6}}})dx &=\int ( \frac{1}{\sqrt {\textcolor{blue}{x^2+x-6}}}).(\textcolor{red}{2x+1})dx\\ &= \int (\frac{1}{\sqrt{\textcolor{blue}{u}}})\textcolor{red}{du} \\ &=\int (\frac{1}{u^{\frac{1}{2}}})du \\ &=\int u^{-\frac{1}{2}} du \\ &= \frac{1}{-\frac{1}{2}+1}u^{-\frac{1}{2}+1}+C \\ &= \frac{1}{\frac{1}{2}}u^{\frac{1}{2}}+C \\ &=2u^{\frac{1}{2}}+C &\to u=x^2+x-6\\ &=2(x^2+x-6)^{\frac{1}{2}}+C \\ &= 2\sqrt{x^2+x-6}+C \end{aligned}$

$\therefore \int (\frac{2x+1}{\sqrt{x^2+x-6}})dx=2\sqrt{x^2+x-6}+C$

Siip, semoga bisa dipahami yah …