Question

Obtain a reduction formula for the following integrals (i) \(\int x^{n} e^{x} d x(n \geq 1)\) (ii) \(\int(\ln x)^{n} d x(n \geq 1)\)

Short answer

Answer: The reduction formulas for the given integrals are: (i) \(I_n = x^{n} e^{x} - n I_{n-1}\) (ii) \(J_n = x (\ln x)^{n} - n J_{n-1}\)

Step-by-step solution

Integration by Parts

Integration by parts formula: \(\int u dv = uv - \int v du\). Choose \(u = x^{n}\) and \(dv = e^{x} dx\). Differentiate u and integrate dv: \(du = n x^{n-1} dx\) \(v = e^{x}\) Apply the formula: \(\int x^{n} e^{x} dx = x^{n} e^{x} - \int n x^{n-1} e^{x} dx\)

Reduction Formula

We can find the reduction formula by comparing the initial integral with the result we got: \(I_n = x^{n} e^{x} - n I_{n-1}\) where \(I_n = \int x^{n} e^{x} dx\) and \(I_{n-1} = \int x^{n-1} e^{x} dx\) For Integral (ii): \(\int(\ln x)^{n} dx\)

Integration by Parts

Integration by parts formula: \(\int u dv = uv - \int v du\). Choose \(u = (\ln x)^{n}\) and \(dv = dx\). Differentiate u and integrate dv: \(du = nx^{1-n} dx\) \(v = x\) Apply the formula: \(\int(\ln x)^{n} dx = x (\ln x)^{n} - n \int (\ln x)^{n-1} dx\)

Reduction Formula

We can find the reduction formula by comparing the initial integral with the result we got: \(J_n = x (\ln x)^{n} - n J_{n-1}\) where \(J_n = \int (\ln x)^{n} dx\) and \(J_{n-1} = \int (\ln x)^{n-1} dx\) In conclusion, the reduction formulas for the given integrals are: (i) \(I_n = x^{n} e^{x} - n I_{n-1}\) (ii) \(J_n = x (\ln x)^{n} - n J_{n-1}\)