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Question Number 77335 by aliesam last updated on 05/Jan/20

Commented by aliesam last updated on 05/Jan/20

Commented by msup trace by abdo last updated on 05/Jan/20

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

Commented by aliesam last updated on 05/Jan/20

$$perfectsirthankyou$$

Commented by mathmax by abdo last updated on 05/Jan/20

$$youarewelcome.$$

Answered by petrochengula last updated on 05/Jan/20

$$=\int \frac{x^3}{x^3+1}dx-\int \frac{1}{x^3+1}dx$$$$=\int (1-\frac{1}{x^3+1})dx-\int \frac{1}{x^3+1}dx$$$$=\int dx-2\int \frac{1}{x^3+1}dx$$$$=x-2\int \frac{1}{(x+1)(x^2-x+1)}dx$$$$=x-2(\int \frac{1}{3(x+1)}dx+\int \frac{2-x}{3(x^2-x+1)}dx)$$$$=x-\frac{2}{3}ln\mid x+1\mid +\frac{2}{3}\int \frac{x-2}{x^2-x+1}dx$$$$Idohopeyouwillbeabletocontinuefromhere$$

Commented by aliesam last updated on 05/Jan/20

$$thankhousir$$

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