What is the integral of (10x^4-4x^2-14x+12) ?

The integral of (10x^4-4x^2-14x+12) is 2x^5-(4x^3)/3-7x^2+12x+C

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integral of (10x^4-4x^2-14x+12)

\int (10 x^{4} - 4 x^{2} - 14 x + 12) d x

2 x^{5} - \frac{4 x ^{3}}{3} - 7 x^{2} + 12 x + C

Solution steps

\int 10 x^{4} - 4 x^{2} - 14 x + 12 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 10 x^{4} d x - \int 4 x^{2} d x - \int 14 x d x + \int 12 d x

\int 10 x^{4} d x = 2 x^{5}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

\int 12 d x = 12 x

= 2 x^{5} - \frac{4 x ^{3}}{3} - 7 x^{2} + 12 x

Add a constant to the solution

= 2 x^{5} - \frac{4 x ^{3}}{3} - 7 x^{2} + 12 x + C

What is the integral of (10x^4-4x^2-14x+12) ?
The integral of (10x^4-4x^2-14x+12) is 2x^5-(4x^3)/3-7x^2+12x+C