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

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

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integral of 7x^4-6x^3+4x^2+C

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

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

Solution steps

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

\int 7 x^{4} d x = \frac{7 x ^{5}}{5}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

\int C d x = C x

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

Add a constant to the solution

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

(2 x^{3} + 5)^{4}

\int (x^{2} + 5 x) cos (x) d x

\frac{d y}{d x} = 0.5 y

\frac{d}{d x} (\frac{8 x ^{2} - 40}{3 x ^{\frac{1}{3}}})

x \to 0 + lim (\frac{∣ 2 x ∣}{x})

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