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Question Number 91990 by Power last updated on 04/May/20

Commented by mathmax by abdo last updated on 04/May/20

$$I=\int \frac{x^3-6}{x^4+6x+8}\Rightarrow I=\frac{1}{4}\int \frac{4x^3-24}{x^4+6x+8}dx\Rightarrow$$$$4I=\int \frac{4x^3+6-30}{x^4+6x+8}dx=ln\mid x^4+6x+8\mid -30\int \frac{dx}{x^4+6x+8}$$$$x^4+6x+8=0therootsare$$$$z_1=1,2402+1,6576i\left(complex\right)$$$$z_2=1,2402-1,6576i\left(complex\right)$$$$z_3=-1,2402+0,5732i\left(complex\right)$$$$z_4=-1,2402-0,5732i\left(complex\right)$$$$let\alpha =1,2402and\beta =0,5732\Rightarrow z_1=\alpha +(1+\beta )i$$$$z_2=\alpha -(1+\beta )i$$$$z_3=-\alpha +\beta iand{z}_4=-\alpha -\beta i\Rightarrow$$$$\int \frac{dx}{x^4+6x+8}=\int \frac{dx}{(x+\alpha -\beta i)(x+\alpha +\beta i)(x-\alpha +(1+\beta )i)(x-\alpha -(1+\beta )i)}$$$$but(x+\alpha -\beta i)(x+\alpha +\beta i)=(x-z_3)(x-{\stackrel{-}{z}}_3)=x^2-2Re\left(z_3\right)x+\mid z_3{\mid }^2$$$$=x^2+2\alpha x+{\alpha }^2+{\beta }^2$$$$(x-z_1)(x-z_2)=(x-z_1)(x-{\stackrel{-}{z}}_1)=x^2-2Re\left(z_1\right)x+\mid z_1{\mid }^2$$$$=x^2-2\alpha x+{\alpha }^2+{(1+\beta )}^2\Rightarrow$$$$\int \frac{dx}{x^4+6x+8}=\int \frac{dx}{(x^2+2\alpha x+{\alpha }^2+{\beta }^2)(x^2-2\alpha x+{\alpha }^2+{(1+\beta )}^2)}$$$$...becontinued....$$

Commented by Power last updated on 04/May/20

$$\mathrm{thanks}$$

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