Berikut 20 soal integrasi (campuran: tak tentu, tentu, substitusi, parsial, trigonometri, rasional) lengkap dengan langkah penyelesaiannya. Saya susun dari yang mudah ke yang lebih menantang. Berikut contoh soalnya:
1) \(\displaystyle \int 5x^3 dx\)
Penyelesaian:
\(\int 5x^3dx = 5\cdot\frac{x^{4}}{4} + C = \frac{5x^4}{4}+C.\)
2) \(\displaystyle \int (3x^2 – 4x + 7),dx\)
Penyelesaian:
\(\int 3x^2dx = x^3\), \(\int -4x,dx = -2x^2\), \(\int 7,dx = 7x\).
Jadi: \(x^3 – 2x^2 + 7x + C.\)
3) \(\displaystyle \int x^{-2},dx\)
Penyelesaian:
Ingat \(\int x^n dx = \frac{x^{n+1}}{n+1}\) untuk \(n\neq -1\).
Dengan \(n=-2\): \(\frac{x^{-1}}{-1} + C = -x^{-1}+C = -\frac{1}{x}+C.\)
4) \(\displaystyle \int \cos(3x),dx\)
Penyelesaian (substitusi sederhana):
\(\int \cos(3x)dx = \frac{1}{3}\sin(3x)+C.\)
5) \(\displaystyle \int \sin^2 x,dx\)
Penyelesaian (identitas):
Gunakan \( \sin^2 x = \frac{1-\cos(2x)}{2}\).
\[\int \sin^2x,dx = \int \frac{1}{2}dx – \int \frac{\cos(2x)}{2}dx= \frac{x}{2} – \frac{1}{4}\sin(2x) + C.\]
6) \(\displaystyle \int x e^{x},dx\)
Penyelesaian (integrasi parsial):
Pilih \(u=x\), \(dv=e^x dx\) → \(du=dx\), \(v=e^x\).
\[\int x e^x dx = x e^x – \int e^x dx = x e^x – e^x + C = e^x(x-1)+C.\]
7) \(\displaystyle \int (2x+1)\sqrt{x},dx\)
Penyelesaian:
Tulis \(\sqrt{x}=x^{1/2}\). Jadi integralnya \(=2x\cdot x^{1/2}+1\cdot x^{1/2}=2x^{3/2}+x^{1/2}\).
Integralkan tiap suku:
\[\int 2x^{3/2}dx = 2\cdot\frac{x^{5/2}}{5/2} = \frac{4}{5}x^{5/2},\quad\int x^{1/2}dx = \frac{2}{3}x^{3/2}.\]
Jadi total: \(\frac{4}{5}x^{5/2}+\frac{2}{3}x^{3/2}+C.\)
8) \(\displaystyle \int \frac{1}{x^2+4},dx\)
Penyelesaian (bentuk arctan):
\(\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)+C\).
Dengan (a=2): \(\frac{1}{2}\arctan\left(\frac{x}{2}\right)+C.\)
9) \(\displaystyle \int \frac{x}{x^2+1},dx\)
Penyelesaian (substitusi):
Ambil \(u=x^2+1\) → du=2x dx → \(x dx=\frac{1}{2}du\).
\[\int \frac{x}{x^2+1}dx = \frac{1}{2}\int \frac{1}{u} du = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln(x^2+1)+C.\]
10) \(\displaystyle \int_0^1 (3x^2+2x),dx\) (integral tentu)
Penyelesaian:
Antiderivatif \(= x^3 + x^2\). Evaluasi dari 0 ke 1: \(1^3+1^2 – (0)=1+1=2.\)
11) \(\displaystyle \int \frac{1}{\sqrt{1-x^2}},dx\)
Penyelesaian:
\(\int \frac{1}{\sqrt{1-x^2}}dx = \arcsin x + C.\)
12) \(\displaystyle \int \tan x,dx\)
Penyelesaian:
\(\tan x = \frac{\sin x}{\cos x}\). Gunakan substitusi \(u=\cos x\), \(du=-\sin x dx\).
\[\int \tan x dx = -\int \frac{1}{u} du = -\ln|u|+C = -\ln|\cos x| + C.\]
(Biasanya ditulis (\ln|\sec x|+C) juga.)
13) \(\displaystyle \int (3x^2-2x+1)e^{x^3 – x^2 + x},dx\)
Penyelesaian (substitusi yang pas):
Perhatikan \(u = x^3 – x^2 + x\) → \(du = (3x^2 – 2x +1)dx\).
Maka integral menjadi \(\int e^u du = e^u + C = e^{x^3 – x^2 + x} + C.\)
14) \(\displaystyle \int \ln x,dx\)
Penyelesaian (integrasi parsial):
Pilih \(u = \ln x\) → \(du = \frac{1}{x}dx\), (dv=dx) → (v=x).
\[\int \ln x dx = x\ln x – \int x\cdot\frac{1}{x}dx = x\ln x – \int 1 dx = x\ln x – x + C.\]
15) \(\displaystyle \int_1^2 \frac{1}{x},dx\)
Penyelesaian:
Antiderivatif (\ln|x|). Evaluasi: (\ln 2 – \ln 1 = \ln 2.)
16) \(\displaystyle \int \frac{dx}{x^2 – 1}\)
Penyelesaian (pecah pembagian parsial):
\[\frac{1}{x^2-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right).\]
Jadi
\[\int \frac{dx}{x^2-1} = \frac{1}{2}\ln|x-1| – \frac{1}{2}\ln|x+1| + C \]\[= \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C.\]
17) \(\displaystyle \int x^2 \cos x,dx\)
Penyelesaian (integrasi parsial dua kali):
Pilih \(u=x^2\), \(dv=\cos x dx\) → \(du=2x dx\), \(v=\sin x\).
\[\int x^2\cos xdx = x^2\sin x – \int 2x\sin x dx.\]
Sekarang integralkan \(\int 2x\sin x dx\). Gunakan parsial lagi: pilih u=2x, \(dv=\sin x dx\) → du=2 dx, \(v=-\cos x\).
\[\int 2x\sin x dx = -2x\cos x + \int 2\cos x dx = -2x\cos x + 2\sin x.\]
Kembali:
\[\int x^2\cos xdx = x^2\sin x – \big(-2x\cos x + 2\sin x\big) + C\] \[= x^2\sin x + 2x\cos x – 2\sin x + C.\]
18) \(\displaystyle \int \frac{dx}{(x+1)^2}\)
Penyelesaian:
Ingat \(\int (x+a)^{-2}dx = -(x+a)^{-1} + C\). Jadi
\[\int \frac{dx}{(x+1)^2} = -\frac{1}{x+1} + C.\]
19) \(\displaystyle \int \frac{x^2}{x^3+1} dx\)
Penyelesaian (substitusi):
Perhatikan \(u=x^3+1\) → \(du=3x^2 dx\) → \(x^2dx=\frac{1}{3}du\).
Sehingga integral jadi \(\frac{1}{3}\int \frac{1}{u} du = \frac{1}{3}\ln|u| + C = \frac{1}{3}\ln(x^3+1) + C.\)
20) \(\displaystyle \int_0^{\pi/2} \sin^3 x dx\)
Penyelesaian (identitas & tentu):
Gunakan \(\sin^3 x = \sin x(1-\cos^2 x)\). Let \(u=\cos x\) → \(du=-\sin x dx\).
\[\int_0^{\pi/2}\sin^3 xdx = \int_0^{\pi/2} \sin x dx – \int_0^{\pi/2} \sin x \cos^2 x dx.\]
Atau langsung substitusi:
\[I=\int_0^{\pi/2}\sin^3 xdx = \int_0^{\pi/2}\sin x(1-\cos^2 x)dx.\]
Dengan \(u=\cos x\), saat \(x=0\Rightarrow u=1\); \(x=\pi/2\Rightarrow u=0\). Maka
\[I = -\int_{1}^{0}(1-u^2)du = \int_{0}^{1}(1-u^2)du= \left[u – \frac{u^3}{3}\right]_0^1 = 1 – \frac{1}{3} = \frac{2}{3}.\]
Demikian pembahasan singkat tentang “Contoh Soal Integral serta Penyelesaian (Langsung & Jelas)“. Semoga bermanfaat.
