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

The integral of (x^4)/4-(x^3)/3-(x^2)/2+x+C is (x^5}{20}-(x^4)/(12)-\frac{x^3)/6+(x^2)/2+Cx+C

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integral of (x^4)/4-(x^3)/3-(x^2)/2+x+C

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

\frac{x ^{5}}{20} - \frac{x ^{4}}{12} - \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + C x + C

Solution steps

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

\int \frac{x ^{4}}{4} d x = \frac{x ^{5}}{20}

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

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

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

\int C d x = C x

= \frac{x ^{5}}{20} - \frac{x ^{4}}{12} - \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + C x

Add a constant to the solution

= \frac{x ^{5}}{20} - \frac{x ^{4}}{12} - \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + C x + C

\int 2 x^{3} d x

\frac{3}{ln ( x )}

x^{2} y^{^{''}} - x y^{^{'}} - 3 y = 0

\frac{x d y}{d x} = \frac{1}{y ^{3}}

\frac{d y}{d x} = (4 + 3 x)^{4}

What is the integral of (x^4)/4-(x^3)/3-(x^2)/2+x+C ?
The integral of (x^4)/4-(x^3)/3-(x^2)/2+x+C is (x^5}{20}-(x^4)/(12)-\frac{x^3)/6+(x^2)/2+Cx+C