What is the integral of (5x^4-3x^4+4x^2-2+6) ?

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

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integral of (5x^4-3x^4+4x^2-2+6)

\int (5 x^{4} - 3 x^{4} + 4 x^{2} - 2 + 6) d x

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

Solution steps

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

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

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

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

\int 2 d x = 2 x

\int 6 d x = 6 x

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

Simplify

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

Add a constant to the solution

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

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