How do you evaluate the indefinite integral $\int (4 x^{5} - 6 x^{3} + 7 x^{2} - 8) d x$ $int (4x^5-6x^3+7x^2-8)dx$ ?

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Answer 1 · 2017-10-22T07:02:30

$\frac{2 x^{6}}{3}$ $(2x^6)/3$ - $\frac{3 x^{4}}{2}$ $(3x^4)/2$ + $\frac{7 x^{3}}{3}$ $(7x^3)/3$ - $8 x$ $8x$

$\int$ $int$ $4 x^{5} d x$ $4x^5 dx$ = $\frac{4 x^{6}}{6}$ $(4x^6)/6$ = $\frac{2 x^{6}}{3}$ $(2x^6)/3$

$\int - 6 x^{3} d x$ $int-6x^3 dx$ = $\frac{- 6 x^{4}}{4}$ $(-6x^4)/4$ = $\frac{- 3 x^{4}}{2}$ $(-3x^4)/2$

$\int 7 x^{2} d x$ $int7x^2 dx$ = $7 \frac{x^{3}}{3}$ $7x^3/3$

$\int - 8 d x$ $int-8 dx$ = $- 8 x$ $-8x$

The above answer uses the formula:

$\int x^{n} d x$ $intx^ndx$ = $\frac{x^{n + 1}}{n + 1}$ $x^(n+1)/(n+1)$

To obtain more reliable answers there are several online calculators available.

Hope my answer helped you.
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How do you evaluate the indefinite integral ∫(4x5−6x3+7x2−8)dxint (4x^5-6x^3+7x^2-8)dx?