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Pattern - 35 of Indefinite Integration

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⋆ Pattern – 35

$I_n=\ \int x^n\ e^{ax}\ dx$

∴ $I_n = \frac{1}{a}\ x^n$ ∙ e^{ax} – $\frac{n}{a} i_{n-1}\$ : n ≥ 1

Here, $I_n=\ \int x^n\ e^{ax}\ dx$

$\therefore\ \ I_n=x^n\int{e^{ax}dx}-\ \int\left(\frac{d}{dx}\ x^n\ \int{e^{ax}\ dx}\right)dx$

$=\ \frac{x^n\times e^{ax}}{a}-\ \int n\ x^{n-1}\times\frac{e^{ax}}{a}\ dx$

$=\ \frac{x^n\ e^{ax}}{a}-\frac{n}{a}\ \int n^{n-1}\ e^{ax}\ dx$

$=\frac{1}{a}\times x^n\times e^{ax}-\frac{n}{a}\ I_{n-1}\ :n\geq1$

1) Evaluate : ∫ $x^3\ e^{4x}\ dx$

Solution : Here, $I_3=\ \int x^3e^{4x}\ dx$

$\therefore\ \ I_3\ =\ \frac{1}{4}x^3e^{4x}\ -\ \frac{3}{4}\ I_2$

$=\ \frac{1}{4\ }\ x^3e^{4x}-\ \frac{3}{4}\left[\frac{1}{4}\ x^2e^{4x}-\frac{1}{2}I_1\right]$

$=\frac{1}{4}x^3\ e^{4x}-\frac{3}{16}x^2e^{4x}\ +\frac{3}{8}\ \left[\frac{1}{4}x\ e^{4x}-\frac{1}{4}I_0\right]$

$=\ \frac{1}{4}\ x^3e^{4x}-\frac{3}{16}x^2e^{4x}+\frac{3}{32}x\ e^{4x}-\frac{3}{32}\times\frac{e^{4x}}{4}+c$

$\left(\because\ I_0=\ \int e^{4x}\ dx=\ \frac{e^{4x}}{4}\right)$