{"id":8186,"date":"2021-11-16T17:59:37","date_gmt":"2021-11-16T12:29:37","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8186"},"modified":"2021-11-16T18:22:20","modified_gmt":"2021-11-16T12:52:20","slug":"integrate-x2-x-1over-x2-1-with-respect-to-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integrate-x2-x-1over-x2-1-with-respect-to-x\/","title":{"rendered":"Integrate \\(x^2 + x – 1\\over x^2 – 1\\) with respect to x."},"content":{"rendered":"
\\(\\int\\) \\(x^2 + x – 1\\over x^2 – 1\\) dx = \\(\\int\\) (\\(x^2 – 1\\over x^2 – 1\\) + \\(x\\over x^2 – 1\\))dx<\/p>\n
= \\(\\int\\) 1 dx + \\(\\int\\) \\(x\\over x^2 – 1\\)) dx<\/p>\n
Let \\(x^2 – 1\\) = t\u00a0 \\(\\implies\\) 2x dx = dt<\/p>\n
= x + \\(\\int\\) \\(dt\\over 2t\\)<\/p>\n
= x + \\(1\\over 2\\) \\(\\int\\) \\(1\\over t\\) dt<\/p>\n
since, integration of \\(1\\over x\\) = log x<\/p>\n
= x + \\(1\\over 2\\) log t + C<\/p>\n
Substituing the value of t here, we get<\/p>\n
= x + \\(log(x^2 – 1)\\over 2\\) + C<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : \\(\\int\\) \\(x^2 + x – 1\\over x^2 – 1\\) dx = \\(\\int\\) (\\(x^2 – 1\\over x^2 – 1\\) + \\(x\\over x^2 – 1\\))dx = \\(\\int\\) 1 dx + \\(\\int\\) \\(x\\over x^2 – 1\\)) dx Let \\(x^2 – 1\\) = t\u00a0 \\(\\implies\\) 2x dx = dt = x + \\(\\int\\) \\(dt\\over 2t\\) = x …<\/p>\n