{"id":4068,"date":"2021-08-15T19:39:13","date_gmt":"2021-08-15T19:39:13","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4068"},"modified":"2021-11-26T01:25:51","modified_gmt":"2021-11-25T19:55:51","slug":"integration-by-partial-fraction","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integration-by-partial-fraction\/","title":{"rendered":"Integration by Partial Fraction Formula"},"content":{"rendered":"
Here you will learn integration by partial fraction formula and integration of irrational functions.<\/p>\n
Let’s begin –<\/p>\n
(i) Integration of Rational Functions<\/strong><\/p>\n\n\n Example : <\/span> Evaluate \\(\\int\\) \\(x\\over {(x-2)(x-5)}\\) dx<\/p>\n Solution : <\/span>We have, \\(\\int\\) \\(x\\over {(x-2)(x-5)}\\) dx Example : <\/span> Evaluate \\(\\int\\) \\(2x\\over {(x^2+1)(x^2+2)}\\) dx<\/p>\n Solution : <\/span>Let I = \\(\\int\\) \\(2x\\over {(x^2+1)(x^2+2)}\\) dx (ii) Integration of Irrational Functions<\/strong><\/p>\n (a) \\(\\int\\) \\(dx\\over {(ax + b)\\sqrt{px+q}}\\) & \\(\\int\\) \\(dx\\over {(ax^2 + bx + c)\\sqrt{px+q}}\\); put px+q = \\(t^2\\)<\/p>\n (b) \\(\\int\\) \\(dx\\over {(ax + b)\\sqrt{px^2+qx+r}}\\); put ax+b = \\(1\\over t\\); \\(\\int\\) \\(dx\\over {(ax^2 + b)\\sqrt{px^2+q}}\\); put x = \\(1\\over t\\) <\/p>\n\n\n\n
\n S.No<\/th>\n form of rational function<\/th>\n form of partial fraction<\/th>\n <\/tr>\n \n 1<\/td>\n \\(px^2+qx+r\\over {(x-a)(x-b)(x-c)}\\)<\/td>\n \\(A\\over {x-a}\\) + \\(B\\over {x-b}\\) + \\(C\\over {x-c}\\)<\/td>\n <\/tr>\n \n 2<\/td>\n \\(px^2+qx+r\\over {{(x-a)}^2(x-b)}\\)<\/td>\n \\(A\\over {x-a}\\) + \\(B\\over {(x-a)}^2\\) + \\(C\\over {x-b}\\)<\/td>\n <\/tr>\n \n 3<\/td>\n \\(px^2+qx+r\\over {(x-a)(x^2+bx+c)}\\)<\/td>\n \\(A\\over {x-a}\\) + \\(Bx+C\\over {x^2+bx+c}\\)<\/td>\n <\/tr>\n <\/tbody><\/table>
\n\n\n\n
\nLet \\(x\\over {(x-2)(x-5)}\\) = \\(A\\over {x-2}\\) + \\(B\\over {x-5}\\)
\n or x = A(x+5) + B(x-2)
\n by comparing the coefficients, we get
\n A = 2\/7 and B = 5\/7 so that
\n \\(\\int\\) \\(x\\over {(x-2)(x-5)}\\) dx = \\(2\\over 7\\) \\(\\int\\)\\(dx\\over x-2\\) + \\(5\\over 7\\) \\(\\int\\)\\(dx\\over x+5\\)
\n = \\(2\\over 7\\) ln|x-2| + \\(5\\over 7\\) ln|x+5| + C
\n <\/p>\n\n\n\n
\nPutting \\(x^2\\) = t and 2xdx = dt, we get
\nI = \\(\\int\\) \\(dt\\over {(t+1)(t+2)}\\)
\nLet \\(1\\over {(t+1)(t+2)}\\) = \\(A\\over t+1\\) + \\(B\\over t+2\\) …….(i)
\n \\(\\implies\\) 1 = A(t+2) + B(t+1) ……..(ii)
\n Putting t = -2 in (ii), we obtain B = -1
\nPutting t = -1 in (ii), we obtain A = 1
\n Putting value of A and B in (i), we get
\n \\(1\\over {(t+1)(t+2)}\\) = \\(1\\over t+1\\) – \\(1\\over t+2\\)
\n I = \\(\\int\\) \\(1\\over {(t+1)(t+2)}\\)
\n\\(\\implies\\) I = \\(\\int\\) \\(1\\over t+1\\)dt – \\(\\int\\) \\(1\\over t+2\\)dt
\n\\(\\implies\\) I = log|t+1| – log|t+2| + C
\n\\(log|x^2+1|\\) – \\(log|x^2+2|\\) + C
\n <\/p>\n\n\n