What is the integral of 7x^5-5x^4+6x^2-14x+3 ?

The integral of 7x^5-5x^4+6x^2-14x+3 is (7x^6)/6-x^5+2x^3-7x^2+3x+C

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integral of 7x^5-5x^4+6x^2-14x+3

\int 7 x^{5} - 5 x^{4} + 6 x^{2} - 14 x + 3 d x

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

Solution steps

\int 7 x^{5} - 5 x^{4} + 6 x^{2} - 14 x + 3 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 5 x^{4} d x + \int 6 x^{2} d x - \int 14 x d x + \int 3 d x

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

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

\int 6 x^{2} d x = 2 x^{3}

\int 14 x d x = 7 x^{2}

\int 3 d x = 3 x

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

Add a constant to the solution

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

(\frac{x ^{2} - 9}{x - 3})^{^{'}}

5 x - x^{2}, 2 x

\int x (6 + x^{2})^{10} d x

\int (x) arctan (x) d x

\frac{d}{d x} (\frac{1}{8 x - 1})

What is the integral of 7x^5-5x^4+6x^2-14x+3 ?
The integral of 7x^5-5x^4+6x^2-14x+3 is (7x^6)/6-x^5+2x^3-7x^2+3x+C