What is the integral of 7x^5+6x^4-3x^3-9x+5 ?

The integral of 7x^5+6x^4-3x^3-9x+5 is (7x^6)/6+(6x^5)/5-(3x^4)/4-(9x^2)/2+5x+C

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integral of 7x^5+6x^4-3x^3-9x+5

\int 7 x^{5} + 6 x^{4} - 3 x^{3} - 9 x + 5 d x

\frac{7 x ^{6}}{6} + \frac{6 x ^{5}}{5} - \frac{3 x ^{4}}{4} - \frac{9 x ^{2}}{2} + 5 x + C

Solution steps

\int 7 x^{5} + 6 x^{4} - 3 x^{3} - 9 x + 5 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 7 x^{5} d x + \int 6 x^{4} d x - \int 3 x^{3} d x - \int 9 x d x + \int 5 d x

\int 7 x^{5} d x = \frac{7 x ^{6}}{6}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

\int 9 x d x = \frac{9 x ^{2}}{2}

\int 5 d x = 5 x

= \frac{7 x ^{6}}{6} + \frac{6 x ^{5}}{5} - \frac{3 x ^{4}}{4} - \frac{9 x ^{2}}{2} + 5 x

Add a constant to the solution

= \frac{7 x ^{6}}{6} + \frac{6 x ^{5}}{5} - \frac{3 x ^{4}}{4} - \frac{9 x ^{2}}{2} + 5 x + C

\int \frac{1}{( x )} d x

\frac{d r}{d θ} = \frac{r ^{2} + 8 r θ}{4 θ ^{2}}

y^{^{''}} - y^{^{'}} + 6 y = 0

\frac{d}{d θ} (\frac{4}{2 cos ( θ ) - sin ( θ )})

f (x) = \frac{8 x}{x ^{2} + 1}, (1, 4)

What is the integral of 7x^5+6x^4-3x^3-9x+5 ?
The integral of 7x^5+6x^4-3x^3-9x+5 is (7x^6)/6+(6x^5)/5-(3x^4)/4-(9x^2)/2+5x+C