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}}}

di mana 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

Karena 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 8). Academic Press, Inc. ISBN 978-0-12-384933-5. LCCN 2014010276.{{cite book}}: Pemeliharaan CS1: Bahasa yang tidak diketahui (link)