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Integral Calculus

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30 Dec 2009 15:24:52 IST

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indefinite integration of log (1+x^3)

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indefinite integration of log (1+x^3)

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akki ~~ unlucky forever ~~

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31 Dec 2009 00:45:17 IST

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$I=\int ln(1+x^3) dx\\ \\ \\applying\ integration\ byparts\\ \\ \\ I=x\ ln(1+x^3)-3\int (\frac{x^3+1-1}{1+x^3})dx\\ \\ \\I=x\ lx(1+x^3)-3x+3\int \frac{dx}{1+x^3}\\ \\ \\let\ I_1 = \int \frac{dx}{1+x^3}=\int \frac{dx}{(x+1)(x^2-x+1)}\\ \\ \\ I_1 = \frac{1}{3} \int \frac{dx}{(x+1)} - \frac{1}{6} \int \frac{2x-1}{x^2-x+1}dx +\frac{1}{2} \int \frac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}}\\ \\ \\ I_1=\frac{ln(x+1)}{3}-\frac{ln(x^2-x+1)}{6}+\frac{1}{\sqrt{3}}tan^{-1}(\frac{2x-1}{\sqrt{3}})\\ \\ \\ so, I=x\ ln(1+x^3)-3x+ln(\frac{x+1}{\sqrt{x^2-x+1}})+\sqrt{3}tan^{-1}(\frac{2x-1}{\sqrt{3}})+C$

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