Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 69894 by Rio Michael last updated on 28/Sep/19

$$\int \frac{2x^5-x}{x^3-2}dx$$

Commented by abdo mathsup 649 cc last updated on 29/Sep/19

$$I=\int \frac{2x^5-x}{x^3-2}dx=\int \frac{2x^2(x^3-2)+4x^2-x}{x^3-2}dx$$$$=\int 2x^{2}dx+\int \frac{4x^2-x}{x^3-2}dx$$$$\int 2x^2=\frac{2}{3}{x}^3+cletdecompose$$$$F\left(x\right)=\frac{4x^2-x}{x^3-2}\Rightarrow F\left(x\right)=\frac{4x^2-x}{(x-\left({}^3\sqrt{2}\right)(x^2+x\left({}^3\sqrt{2}\right)+{\left({}^3\sqrt{2}\right)}^2)}$$$$letput{}^3\sqrt{2}=\alpha \Rightarrow F\left(x\right)=\frac{4x^2-x}{(x-\alpha )(x^2+\alpha x+{\alpha }^2)}$$$$=\frac{a}{x-\alpha }+\frac{bx+c}{x^2+\alpha x+{\alpha }^2}$$$$a=\frac{4{\alpha }^2-\alpha }{3{\alpha }^2}=\frac{4\alpha -1}{3\alpha }=\frac{4}{3}-\frac{1}{3\alpha }$$$${lim}_{x-\to +\mathrm{\infty }}xF\left(x\right)=4=a+b\Rightarrow b=4-a$$$$=4-(\frac{4}{3}-\frac{1}{3\alpha })=\frac{8}{3}+\frac{1}{3\alpha }$$$$F\left(0\right)=0=-\frac{a}{\alpha }+\frac{c}{{\alpha }^2}=\frac{c-\alpha a}{{\alpha }^2}\Rightarrow c=\alpha a=\frac{4\alpha }{3}-\frac{1}{3}$$$$\int F\left(x\right)dx=aln\mid x-\alpha \mid +\frac{1}{2}\int \frac{2bx+2c}{x^2+\alpha x+{\alpha }^2}dx$$$$=aln\mid x-\alpha \mid +\frac{b}{2}\int \frac{2x+\alpha -\alpha +2c}{x^2+\alpha x+{\alpha }^2}dx$$$$=aln\mid x-\alpha \mid +\frac{b}{2}ln(x^2+\alpha x+{\alpha }^2)+\frac{(2c-\alpha )b}{2}\int \frac{dx}{x^2+\alpha x+{\alpha }^2}$$$$and\int \frac{dx}{x^2+\alpha x+{\alpha }^2}=\int \frac{dx}{x^2+2\frac{\alpha }{2}x+\frac{{\alpha }^2}{4}+{\alpha }^2-\frac{{\alpha }^2}{4}}$$$$=\int \frac{dx}{{(x+\frac{\alpha }{2})}^2+\frac{3{\alpha }^2}{4}}{=}_{x+\frac{\alpha }{2}=\frac{\sqrt{3}}{2}u}$$$$=\frac{4}{3{\alpha }^2}\int \frac{1}{1+u^2}\frac{\sqrt{3}}{2}du=\frac{2}{\sqrt{3}{\alpha }^2}arctan\left(\frac{2x+\alpha }{\sqrt{3}}\right)+c\Rightarrow$$$$I=\frac{2}{3}{x}^3+aln\mid x-\alpha \mid +\frac{b}{2}ln(x^2+\alpha x+{\alpha }^2)$$$$+\frac{(2c-\alpha )b}{\sqrt{3}{\alpha }^2}arctan\left(\frac{2x+\alpha }{\sqrt{3}}\right)+Cwith\alpha {=}^3\sqrt{2}$$

Commented by Rio Michael last updated on 29/Sep/19

$$thankssir$$

Commented by mathmax by abdo last updated on 29/Sep/19

$$youarewelcome.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com