Evaluate \int x^3(\ln(4x))^2 dx by taking advantage of the reduction formula.

Question:

Evaluate {eq}\int x^3(\ln(4x))^2 dx {/eq} by taking advantage of the reduction formula.

Integration


Reduction Formula: Reduction formulas are the formulas used to simplify integrands to a more easily integrated form. The reduction can be obtained by using any of several methods such as integration by parts, integration by substitution, or integration by partial fractions.


{eq}\displaystyle{\int x^n\ln^m(x) \ dx = \frac{x^{n+1}\ln ^m(x)}{n+ 1} - \frac{m}{n + 1}\int x^n\ln^{m -1}(x) \ dx\\} {/eq}

Answer and Explanation:


{eq}\displaystyle{\int x^3(\ln(4x))^2 dx\\[10pt] \text{Using Reduction formula}\quad\int x^n\ln^m(x) \ dx = \frac{x^{n+1}\ln ^m(x)}{n+ 1} - \frac{m}{n + 1}\int x^n\ln^{m -1}(x) \ dx\\[10pt] \text{with n = 3 and m =2}\\[10pt] \int x^3(\ln(4x))^2 dx = \frac{x^4\ln^2(4x)}{4} - \frac{2}{4}\int x^3\ln (4x) \ dx\\[10pt] \int x^3(\ln(4x))^2 dx = \frac{x^4\ln^2(4x)}{4} - \frac{1}{2}\int x^3\ln (4x) \ dx\hspace{35pt}(1)\\[10pt] \text{ We solve the integral part of (1) separately}\\[10pt] \text{Let I = }\int x^3\ln(4x) \ dx\\[10pt] \text{ Applying integration by parts}\int uv\hspace{5pt}dx=u\int v\hspace{5pt}dx-\int \frac{d}{dx}(u)(\int v\hspace{5pt}dx)dx\\[10pt] \text{ Here, we choose first function that allows differentation to be simpler, so we choose}\ln (4x)\quad\text{to be the first function}\\[10pt] \text{Let u = }\ ln (4x)\quad\text{and v = }x^3\\[10pt] I = \ln (4x)\int x^3 \ dx - \int\frac{d}{dx}\ln (4x)\left(\int x^3 \ dx\right) \ dx\\[10pt] = \frac{x^4\ln (4x)}{4} - \int\frac{1}{4x}\cdot 4\frac{x^4}{4} \ dx\\[10pt] = \frac{x^4\ln (4x)}{4} - \frac{1}{4}\int x^3 \ dx\\[10pt] = \frac{x^4\ln (4x)}{4} - \frac{1}{4}\cdot \frac{x^4}{4} + C\\[10pt] = \frac{x^4\ln (4x)}{4} - \frac{x^4}{16} + C\\[10pt] = \frac{4x^4\ln(4x) - x^4}{16} + C\quad\quad\quad\quad\quad\quad\quad(2)\hspace{50pt}\text{(Taking L.C.M.)}\\[10pt] \text{ Putting (2) in (1) we get}\\[10pt] = \frac{x^4\ln^2(4x)}{4} - \frac{1}{2}\frac{4x^4\ln (4x) - x^4}{16}\\[10pt] = \frac{x^4\ln^2(4x)}{4} - \frac{4x^4\ln (4x) - x^4}{32}\\[10pt] = \frac{8x^4\ln^2(4x) - 4x^4\ln (4x) +x^4}{32}\hspace{90pt}\text{(Taking L.C.M.)}\\[10pt] \boxed{ = \frac{x^4\left(8\ln^2(4x) - 4\ln(4x) + 1\right)}{32} + C}\\} {/eq}


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