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Dalam matematika, integral kuadratik adalah integral dengan bentuk umum
![{\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36fb55bbe1234234b2b82bf70478a963c69bdd50)
dimana nilai
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0bd91de5ac6e042b81c29deb74a5563adba11d9)
Diasumsikan nilai diskriminan
. Dalam kasus ini, didefinisikan variabel pembantu
![{\displaystyle 2ax+b=t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e604a2394ea3afea8df74086906bb520d18977c5)
![{\displaystyle b^{2}-4ac=k^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/950b55b3726e4a60eaea01866b1d6dda581fff12)
yang mengakibatkan
dan
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4563234133edef2fdd62347abca0a923c27c7667)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15be55a91d088fe52b3bfa3b8fedf5995c985986)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8062dff434153b8394ebe81523ebeae3b70e6fcd)
Pada kasus ini, informasi nilai
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0da90348a78f95b39ab20654b9fb9d9a9c19f9a5)
Dengan menggunakan substitusi
(yang berarti
), 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28994c38285ed84fe58d40a52cc946f01a4f975f)
Dikarenakan nilai diskriminan
, 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43c1873c8f75d8561d0abda482fa172cf544d8e1)
bernilai positif, sehingga akan digunakan substitusi
![{\displaystyle 2ax+b={\sqrt {4ac-b^{2}}}\tan t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ba9be1b62785cd50d68c72370589cd36ddaf4fd)
![{\displaystyle 2a\,dx={\sqrt {4ac-b^{2}}}\sec ^{2}t\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/342dfef5ffe990edaad48b5a2dfdb30fe0f654d4)
(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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a808bf6d56aca7ea9bf7d436cb16bbcef1b2ac53)
- 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.