How do you evaluate the integral $\int x^{10} - 6 x^{5} + 2 x^{3} d x$ $intx^10-6x^5+2x^3 dx$ ?

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Answer 1 · 2018-06-28T20:00:47

$\frac{x^{11}}{11} - x^{6} + \frac{x^{4}}{2} + c$ $x^11/11-x^6+x^4/2+c$

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c, x \neq - 1$ $intx^ndx=x^(n+1)/(n+1)+c,x!=-1$

$\int (x^{10} - 6 x^{5} + 2 x^{3}) d x$ $int(x^10-6x^5+2x^3)dx$

$= \frac{x^{11}}{11} - \frac{6 x^{6}}{6} + \frac{2 x^{4}}{4} + c$ $=x^11/11-(6x^6)/6+(2x^4)/4+c$

cancelling down we have

$\frac{x^{11}}{11} - x^{6} + \frac{x^{4}}{2} + c$ $x^11/11-x^6+x^4/2+c$

How do you evaluate the integral intx^10-6x^5+2x^3 dx∫x10−6x5+2x3dxintx^10-6x^5+2x^3 dx?