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integrate x/((1+(x)^3))
Note that x³+1 = (x+1)(x²-x+1). You''ll wind up with a partial fraction decomposition with two logarithmic integrals, plus a trig integral. Start as follows: x/(x³+1) = A/(x+1) + (Bx+C)/(x²-x+1)
and find the values of A,B,C....
and solve.
I get 1/6(ln(x^2 -x +1) -1/3(ln(x+1)) + arctan(2x-1/root3)/root3
Please approve the answer.... :)
apply the formula1+x^3=(1+x)(1+x^2-x)and then apply partial fractions.and then proceed.
x/(1+ x^3)=1/x^2 / ( 1/x^3 + 1)let 1/x^3 + 1 = tthen -3/x^2 dx = dtso the integrl reduces to int( -1/3 * [dt/t]
=-1/3*ln|t| + c=-1/3 * ln | 1/x^3 + 1 | + c
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