1708\hspace{-16}\int\frac{x^3}{(1+x^3)^2}dx$\\\\\\ $\frac{1}{3}.\int x.\frac{3x^2}{(1+x^3)^2}dx$\\\\ Now Using Integration by parts, We Get\\\\ $-\frac{1}{3}.x.\frac{1}{(1+x^3)}+\frac{1}{3}.\int\frac{1}{1+x^3}dx$\\\\
\hspace{-16}(1)\;\; \int\frac{x^3}{(1+x^3)^2}dx\\\\\\ (2)\;\; \int\frac{2012x-1}{e^{2012x}-2011x}dx
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3 Answers
711.
Again it's getting lengthy. A very complicated use of partial fraction has to be made I guess.
I could complete it till here:
I = ∫dx(1+x)(1-x+x2) -∫dx(1+x)2(1-x+x2)2
I1 I2
I could evaluate I1 by Partial fractions. But Didn't try for I2.
Anyone to complete this or via other method.
1708