FP3 June 2014 Q9
9. \[I_n = \int (x^2 + 1)^{-n}\,\mathrm{d}x, \qquad n > 0\]
(a) Show that, for \(n > 0\) \[I_{n+1} = \frac{x(x^2 + 1)^{-n}}{2n} + \frac{2n - 1}{2n}I_n\] (5)
(b) Find \(I_2\) (3)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int (x^2 + 1)^{-n}\,\mathrm{d}x = x(x^2 + 1)^{-n} + \int xn(x^2 + 1)^{-n-1}2x\,\mathrm{d}x\) M1: Integration by parts in the correct direction A1: Correct expression (If the parts formula is not quoted and the expression is wrong, score M0A0) | M1A1 |
| \(= x(x^2 + 1)^{-n} + 2n\displaystyle\int x^2(x^2 + 1)^{-n-1}\,\mathrm{d}x\) | |
| \(= x(x^2 + 1)^{-n} + 2n\displaystyle\int (x^2 + 1)^{-n} - (x^2 + 1)^{-n-1}\,\mathrm{d}x\) Use of \(x^2 = x^2 + 1 - 1\) or equivalent. Dependent on the previous method mark. | dM1 |
| \(I_n = x(x^2 + 1)^{-n} + 2nI_n - 2nI_{n+1}\) Correctly replaces \(\displaystyle\int (x^2 + 1)^{-n}\,\mathrm{d}x\) and \(\displaystyle\int (x^2 + 1)^{-n-1}\,\mathrm{d}x\) by \(I_n\) and \(I_{n+1}\). Dependent on both previous method marks. | ddM1 |
| \(I_{n+1} = \dfrac{x(x^2 + 1)^{-n}}{2n} + \dfrac{2n - 1}{2n}I_n\) Correct completion to the printed answer with no errors. | A1cso |
| (5) |
| Scheme | Marks |
|---|---|
| \(I_2 = \dfrac{x(x^2 + 1)^{-1}}{2} + \dfrac{1}{2}I_1\) Correct application of the given reduction formula using \(n = 1\) only | M1 |
| \(I_1 = \displaystyle\int \frac{\mathrm{d}x}{x^2 + 1} = \arctan x\,(+C)\) \(I_1 = k\arctan x\) (must be \(x\) and not just for arctan) | M1 |
| \(I_2 = \dfrac{x}{2(x^2 + 1)} + \dfrac{1}{2}\arctan x\,(+C)\) Cao (constant not needed) | A1 |
| (3) | |
| (8 marks) |