integral (x/(3x^2+2x-1))dx Solution integral x/(-1+2 x+3 x^2) dx x/(3 x^2+2 x-1)

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integral x/(-1+2 x+3 x^2) dx x/(3 x^2+2 x-1) => (6 x+2)/(6 (3 x^2+2 x-1)) -1/(3 (3 x^2+2 x-1)) = integral ((6 x+2)/(6 (3 x^2+2 x-1)) -1/(3 (3 x^2+2 x-1))) dx = 1/6 integral (6 x+2)/(3 x^2+2 x-1) dx-1/3 integral 1/(3 x^2+2 x-1) dx integrand (6 x+2)/(3 x^2+2 x-1), u = 3 x^2+2 x-1 and du = 6 x+2 dx: = 1/6 integral 1/u du-1/3 integral 1/(3 x^2+2 x-1) dx For integrand 1/(3 x^2+2 x-1), = 1/6 integral 1/u du-1/3 integral 1/((sqrt(3) x+1/sqrt(3))^2-4/3) dx For integrand 1/((sqrt(3) x+1/sqrt(3))^2-4/3) s = sqrt(3) x+1/sqrt(3) and ds = sqrt(3) dx: = 1/6 integral 1/u du-1/(3 sqrt(3)) integral 1/(s^2-4/3) ds The integral of 1/(s^2-4/3) is -1/2 sqrt(3) tanh^(-1)((sqrt(3) s)/2): = 1/6 tanh^(-1)((sqrt(3) s)/2)+1/6 integral 1/u du integral of 1/u is log(u): = 1/6 tanh^(-1)((sqrt(3) s)/2)+(log(u))/6+constant s = sqrt(3) x+1/sqrt(3): = (log(u))/6+1/6 tanh^(-1)((3 x)/2+1/2)+constant u = 3 x^2+2 x-1: = 1/6 log(3 x^2+2 x-1)+1/6 tanh^(-1)((3 x)/2+1/2)+constant = 1/6 (log(3 x^2+2 x-1)+tanh^(-1)((3 x)/2+1/2))+constant = 1/12 (log(1-3 x)+3 log(x+1))+constant

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