What is the integral of (15x^4+7x-1) ?

The integral of (15x^4+7x-1) is 3x^5+(7x^2)/2-x+C

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integral of (15x^4+7x-1)

\int (15 x^{4} + 7 x - 1) d x

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

Solution steps

\int 15 x^{4} + 7 x - 1 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 15 x^{4} d x + \int 7 x d x - \int 1 d x

\int 15 x^{4} d x = 3 x^{5}

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

\int 1 d x = x

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

Add a constant to the solution

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

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What is the integral of (15x^4+7x-1) ?
The integral of (15x^4+7x-1) is 3x^5+(7x^2)/2-x+C