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Pattern - 36 of Indefinite Integration

February 20, 2021
Category: Pattern - 36 of Indefinite Integration

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Pattern 36 :

$\ I_n=\ \int\left(\log{x}\right)^n\ dx, then$

$I_n = \left(\log{x}\right)^n- n I_{n-1}$ : n ≥ 1

Here, $I_n=\ \int\left(\log{x}\right)^n\ dx$

$\therefore\ \ I_n=\ \int \left(\log{x}\right)^n \times 1 dx$

$=\ \left(\log{x}\right)^n\ \int{dx-\ }\int\left(\frac{d}{dx}\left(\log{x}\right)^n\ \int dx\right)\ dx$

$=x\times \left(\log{x}\right)^n-\ \int n\ \left(\log{x}\right)^{n-1}\times\frac{1}{x}\times x\ dx$

$=x\ \left(\log{x}\right)^n-n\ \int\left(\log{x}\right)^{n-1}\ dx$

$=x\left(\log{x}\right)^n-n\ I_{n-1}\ : n \geq 1$

1) Evaluate : $\int\left(\log{x}\right)^5\ dx$

Solution : Let, $I_5=\ \int\left(\log{x}\right)^5\ dx$

∴ $I_5 = x\left(\log{x}\right)^5-5\ I_4$

$=x\ \left(\log{x}\right)^5-5\ \left[x\ \left(\log{x}\right)^4-4\ I_3\right]$

$=x\ \left(\log{x}\right)^5-5x\ \left(\log{x}\right)^4+20\left[x\left(\log{x}\right)^3-3I_2\right]$

$=x\left(\log{x}\right)^5-5x\left(\log{x}\right)^4+20\ x\left(\log{x}\right)^3-60\ \left[x\left(\log{x}\right)^2-2I_1\right]$

$=x\left(\log{x}\right)^5-5x\ \left(\log{x}\right)^4+20x\ \left(\log{x}\right)^3-60x\ \left(\log{x}\right)^2+120\ \left[x\log{x}-I_0\right]$

$=x\left(\log{x}\right)^5-5x\ \left(\log{x}\right)^4+20\ x\left(\log{x}\right)^3-60x\ \left(\log{x}\right)^2 + 120x\ log \times - 120$ x + c ( ∵ $I_0$ = ∫ 1 dx = x )

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Pattern - 36 of Indefinite Integration

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