- Source: Integral kuadratik
Dalam matematika, integral kuadratik adalah integral dengan bentuk umum
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}}
dimana nilai
a
≠
0
{\displaystyle a\neq 0}
. Integral di atas dapat diselesaikan dengan melengkapkan kuadrat sempurna pada bagian penyebut, yaitu sebagai berikut
∫
d
x
a
x
2
+
b
x
+
c
=
∫
4
a
4
a
2
x
2
+
4
a
b
x
+
4
a
c
d
x
=
∫
4
a
(
2
a
x
)
2
+
(
2
a
x
)
(
b
)
+
b
2
−
b
2
+
4
a
c
d
x
=
∫
4
a
(
2
a
x
+
b
)
2
−
(
b
2
−
4
a
c
)
d
x
{\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{4a^{2}x^{2}+4abx+4ac}}\,dx\\&=\int {\frac {4a}{\left(2ax\right)^{2}+(2ax)(b)+b^{2}-b^{2}+4ac}}\,dx\\&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\\end{aligned}}}
Kasus Diskriminan Positif
Diasumsikan nilai diskriminan
b
2
−
4
a
c
>
0
{\displaystyle b^{2}-4ac>0}
. Dalam kasus ini, didefinisikan variabel pembantu
2
a
x
+
b
=
t
{\displaystyle 2ax+b=t}
b
2
−
4
a
c
=
k
2
{\displaystyle b^{2}-4ac=k^{2}}
yang mengakibatkan
2
a
d
x
=
d
t
{\displaystyle 2a\,dx=dt}
dan
k
=
b
2
−
4
a
c
{\displaystyle k={\sqrt {b^{2}-4ac}}}
. Dari sini, integral kuadratiknya menjadi
∫
d
x
a
x
2
+
b
x
+
c
=
∫
4
a
(
2
a
x
+
b
)
2
−
(
b
2
−
4
a
c
)
d
x
=
∫
2
t
2
−
k
2
d
t
=
∫
2
(
t
−
k
)
(
t
+
k
)
d
t
{\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\&=\int {\frac {2}{t^{2}-k^{2}}}\,dt\\&=\int {\frac {2}{(t-k)(t+k)}}\,dt\end{aligned}}}
Dengan menggunakan teknik dekomposisi pecahan parsial, perhatikan bahwa
2
(
t
−
k
)
(
t
−
k
)
=
1
k
(
1
t
−
k
−
1
t
+
k
)
{\displaystyle {\frac {2}{(t-k)(t-k)}}={\frac {1}{k}}\left({\frac {1}{t-k}}-{\frac {1}{t+k}}\right)}
Sehingga diperoleh
∫
d
x
a
x
2
+
b
x
+
c
=
∫
2
(
t
−
k
)
(
t
+
k
)
d
t
=
∫
1
k
(
1
t
−
k
−
1
t
+
k
)
d
t
=
1
b
2
−
4
a
c
(
ln
|
t
−
k
|
−
ln
|
t
+
k
|
)
+
konstanta
=
1
b
2
−
4
a
c
ln
|
t
−
k
t
+
k
|
+
konstanta
=
1
b
2
−
4
a
c
ln
|
2
a
x
+
b
−
b
2
−
4
a
c
2
a
x
+
b
+
b
2
−
4
a
c
|
+
konstanta
{\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {2}{(t-k)(t+k)}}\,dt\\&=\int {\frac {1}{k}}\left({\frac {1}{t-k}}-{\frac {1}{t+k}}\right)\,dt\\&={\frac {1}{\sqrt {b^{2}-4ac}}}\left(\ln \left|t-k\right|-\ln \left|t+k\right|\right)+{\text{konstanta}}\\&={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {t-k}{t+k}}\right|+{\text{konstanta}}\\&={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+{\text{konstanta}}\end{aligned}}}
Kasus Diskriminan Nol
Pada kasus ini, informasi nilai
b
2
−
4
a
c
=
0
{\displaystyle b^{2}-4ac=0}
akan mempermudah pengerjaan integral kuadratiknya, karena
∫
d
x
a
x
2
+
b
x
+
c
=
∫
4
a
(
2
a
x
+
b
)
2
−
(
b
2
−
4
a
c
)
d
x
=
∫
4
a
(
2
a
x
+
b
)
2
d
x
{\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\&=\int {\frac {4a}{\left(2ax+b\right)^{2}}}\,dx\end{aligned}}}
Dengan menggunakan substitusi
t
=
2
a
x
+
b
{\displaystyle t=2ax+b}
(yang berarti
d
t
=
2
a
d
x
{\displaystyle dt=2a\,dx}
), maka
∫
d
x
a
x
2
+
b
x
+
c
=
∫
4
a
(
2
a
x
+
b
)
2
d
x
=
∫
2
t
2
d
t
=
−
2
t
+
konstanta
=
−
2
2
a
x
+
b
+
konstanta
{\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}}}\,dx\\&=\int {\frac {2}{t^{2}}}\,dt\\&=-{\frac {2}{t}}+{\text{konstanta}}\\&=-{\frac {2}{2ax+b}}+{\text{konstanta}}\end{aligned}}}
Kasus Diskriminan Negatif
Dikarenakan nilai diskriminan
b
2
−
4
a
c
<
0
{\displaystyle b^{2}-4ac<0}
, maka suku kedua pada bagian penyebut dari
∫
d
x
a
x
2
+
b
x
+
c
=
∫
4
a
(
2
a
x
+
b
)
2
−
(
b
2
−
4
a
c
)
d
x
=
∫
4
a
(
2
a
x
+
b
)
2
+
(
4
a
c
−
b
2
)
d
x
{\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\&=\int {\frac {4a}{\left(2ax+b\right)^{2}+\left(4ac-b^{2}\right)}}\,dx\end{aligned}}}
bernilai positif, sehingga akan digunakan substitusi
2
a
x
+
b
=
4
a
c
−
b
2
tan
t
{\displaystyle 2ax+b={\sqrt {4ac-b^{2}}}\tan t}
2
a
d
x
=
4
a
c
−
b
2
sec
2
t
d
t
{\displaystyle 2a\,dx={\sqrt {4ac-b^{2}}}\sec ^{2}t\,dt}
tan
2
t
+
1
=
sec
2
t
{\displaystyle \tan ^{2}t+1=\sec ^{2}t}
(lihat identitas Pythagoras)
Akibatnya,
∫
d
x
a
x
2
+
b
x
+
c
=
∫
4
a
(
2
a
x
+
b
)
2
+
(
4
a
c
−
b
2
)
d
x
=
∫
2
(
4
a
c
−
b
2
tan
t
)
2
+
(
4
a
c
−
b
2
)
⋅
4
a
c
−
b
2
sec
2
t
d
t
=
∫
2
(
4
a
c
−
b
2
)
(
tan
2
t
+
1
)
⋅
4
a
c
−
b
2
sec
2
t
d
t
=
2
4
a
c
−
b
2
∫
sec
2
t
sec
2
t
d
t
=
2
4
a
c
−
b
2
t
+
konstanta
=
2
4
a
c
−
b
2
arctan
(
2
a
x
+
b
4
a
c
−
b
2
)
+
konstanta
.
{\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}+\left(4ac-b^{2}\right)}}\,dx\\&=\int {\frac {2}{\left({\sqrt {4ac-b^{2}}}\tan t\right)^{2}+\left(4ac-b^{2}\right)}}\cdot {\sqrt {4ac-b^{2}}}\sec ^{2}t\,dt\\&=\int {\frac {2}{\left(4ac-b^{2}\right)\left(\tan ^{2}t+1\right)}}\cdot {\sqrt {4ac-b^{2}}}\sec ^{2}t\,dt\\&={\frac {2}{\sqrt {4ac-b^{2}}}}\int {\dfrac {\sec ^{2}t}{\sec ^{2}t}}\,dt\\&={\frac {2}{\sqrt {4ac-b^{2}}}}t+{\text{konstanta}}\\&={\frac {2}{\sqrt {4ac-b^{2}}}}\arctan \left({\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\right)+{\text{konstanta}}.\end{aligned}}}
Referensi
Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Web Resource, wherein the following is referenced:
Gradshteyn, Izrail Solomonovich; Ryzhik, Iosif Moiseevich; Geronimus, Yuri Veniaminovich; Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo, ed. Table of Integrals, Series, and Products (dalam bahasa English). Diterjemahkan oleh Scripta Technica, Inc. (edisi ke-8). Academic Press, Inc. ISBN 978-0-12-384933-5. LCCN 2014010276. Pemeliharaan CS1: Bahasa yang tidak diketahui (link)
Kata Kunci Pencarian:
- Integral
- Integral kuadratik
- Rata-rata kuadrat
- Pangkat dua
- Luas
- Derajat kebebasan (fisika dan kimia)
- Kaidah Simpson
- Kalkulus diferensial
- Integrasi cakram
- Bilangan prima
