Integral kuadratik

Dalam matematika, integral kuadratik adalah integral dengan bentuk umum

${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}}$

di mana nilai ${\displaystyle a\neq 0}$. Integral di atas dapat diselesaikan dengan melengkapkan kuadrat sempurna pada bagian penyebut, yaitu sebagai berikut

${\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{4a^{2}x^{2}+4abx+4ac}}\,dx\\&=\int {\frac {4a}{\left(2ax\right)^{2}+(2ax)(b)+b^{2}-b^{2}+4ac}}\,dx\\&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\\end{aligned}}}$

Diasumsikan nilai diskriminan ${\displaystyle b^{2}-4ac>0}$. Dalam kasus ini, didefinisikan variabel pembantu

• ${\displaystyle 2ax+b=t}$
• ${\displaystyle b^{2}-4ac=k^{2}}$

yang mengakibatkan ${\displaystyle 2a\,dx=dt}$ dan ${\displaystyle k={\sqrt {b^{2}-4ac}}}$. Dari sini, integral kuadratiknya menjadi

${\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\&=\int {\frac {2}{t^{2}-k^{2}}}\,dt\\&=\int {\frac {2}{(t-k)(t+k)}}\,dt\end{aligned}}}$

Dengan menggunakan teknik dekomposisi pecahan parsial, perhatikan bahwa

${\displaystyle {\frac {2}{(t-k)(t-k)}}={\frac {1}{k}}\left({\frac {1}{t-k}}-{\frac {1}{t+k}}\right)}$

Sehingga diperoleh

${\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {2}{(t-k)(t+k)}}\,dt\\&=\int {\frac {1}{k}}\left({\frac {1}{t-k}}-{\frac {1}{t+k}}\right)\,dt\\&={\frac {1}{\sqrt {b^{2}-4ac}}}\left(\ln \left|t-k\right|-\ln \left|t+k\right|\right)+{\text{konstanta}}\\&={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {t-k}{t+k}}\right|+{\text{konstanta}}\\&={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+{\text{konstanta}}\end{aligned}}}$

Pada kasus ini, informasi nilai ${\displaystyle b^{2}-4ac=0}$ akan mempermudah pengerjaan integral kuadratiknya, karena

${\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\&=\int {\frac {4a}{\left(2ax+b\right)^{2}}}\,dx\end{aligned}}}$

Dengan menggunakan substitusi ${\displaystyle t=2ax+b}$ (yang berarti ${\displaystyle dt=2a\,dx}$), maka

${\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}}}\,dx\\&=\int {\frac {2}{t^{2}}}\,dt\\&=-{\frac {2}{t}}+{\text{konstanta}}\\&=-{\frac {2}{2ax+b}}+{\text{konstanta}}\end{aligned}}}$

Dikarenakan nilai diskriminan ${\displaystyle b^{2}-4ac<0}$, maka suku kedua pada bagian penyebut dari

${\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}-\left(b^{2}-4ac\right)}}\,dx\\&=\int {\frac {4a}{\left(2ax+b\right)^{2}+\left(4ac-b^{2}\right)}}\,dx\end{aligned}}}$

bernilai positif, sehingga akan digunakan substitusi

• ${\displaystyle 2ax+b={\sqrt {4ac-b^{2}}}\tan t}$
• ${\displaystyle 2a\,dx={\sqrt {4ac-b^{2}}}\sec ^{2}t\,dt}$
• ${\displaystyle \tan ^{2}t+1=\sec ^{2}t}$ (lihat identitas Pythagoras)

Akibatnya,

${\displaystyle {\begin{aligned}\int {\frac {dx}{ax^{2}+bx+c}}&=\int {\frac {4a}{\left(2ax+b\right)^{2}+\left(4ac-b^{2}\right)}}\,dx\\&=\int {\frac {2}{\left({\sqrt {4ac-b^{2}}}\tan t\right)^{2}+\left(4ac-b^{2}\right)}}\cdot {\sqrt {4ac-b^{2}}}\sec ^{2}t\,dt\\&=\int {\frac {2}{\left(4ac-b^{2}\right)\left(\tan ^{2}t+1\right)}}\cdot {\sqrt {4ac-b^{2}}}\sec ^{2}t\,dt\\&={\frac {2}{\sqrt {4ac-b^{2}}}}\int {\dfrac {\sec ^{2}t}{\sec ^{2}t}}\,dt\\&={\frac {2}{\sqrt {4ac-b^{2}}}}t+{\text{konstanta}}\\&={\frac {2}{\sqrt {4ac-b^{2}}}}\arctan \left({\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\right)+{\text{konstanta}}.\end{aligned}}}$

• 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. Pemeliharaan CS1: Bahasa yang tidak diketahui (link)