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Question Number 77960 by jagoll last updated on 12/Jan/20

$$\int \frac{2x^3-1}{x^4+x}dx?$$

Commented by john santu last updated on 12/Jan/20

$$wedivideby{x}^2$$$$\int \frac{2x-\frac{1}{x^2}}{x^2+\frac{1}{x}}dx$$$$nowbyusingintegration$$$$substitution‘‘u‘‘$$$$letu=x^2+\frac{1}{x}\Rightarrow du=2x-\frac{1}{x^2}dx$$$$\int \frac{du}{u}=ln\mid u\mid +c=ln\mid x^2+\frac{1}{x}\mid +c$$

Commented by jagoll last updated on 12/Jan/20

$$thankyou$$

Commented by mathmax by abdo last updated on 13/Jan/20

$$letdecomposeF\left(x\right)=\frac{2x^3-1}{x^4+x}\Rightarrow F\left(x\right)=\frac{2x^3-1}{x(x^3+1)}$$$$=\frac{2x^3-1}{x(x+1)(x^2-x+1)}=\frac{a}{x}+\frac{b}{x+1}+\frac{cx+d}{x^2-x+1}$$$$a=xF\left(x\right){\mid }_{x=0}=-1$$$$b=(x+1)F\left(x\right){\mid }_{x=-1}=\frac{-3}{-3}=1\Rightarrow F\left(x\right)=-\frac{1}{x}+\frac{1}{x+1}+\frac{cx+d}{x^2-x+1}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=2=c\Rightarrow F\left(x\right)=-\frac{1}{x}+\frac{1}{x+1}+\frac{2x+d}{x^2-x+1}$$$$F\left(1\right)=\frac{1}{2}=-1+\frac{1}{2}+2+d\Rightarrow d=-1\Rightarrow$$$$F\left(x\right)=-\frac{1}{x}+\frac{1}{x+1}+\frac{2x-1}{x^2-x+1}\Rightarrow \int F\left(x\right)dx=-ln\mid x\mid +ln\mid x+1\mid +ln(x^2-x+1)+c$$$$$$$$=ln\mid x^3+1\mid -ln\mid x\mid +c=ln\mid \frac{x^3+1}{x}\mid +C$$$$=ln\mid x^2+\frac{1}{x}\mid +C$$$$$$

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