What is the integral of (u^5-3u^4-u^2+2/7) ?

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

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integral of (u^5-3u^4-u^2+2/7)

\int (u^{5} - 3 u^{4} - u^{2} + \frac{2}{7}) d u

\frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u + C

Solution steps

\int u^{5} - 3 u^{4} - u^{2} + \frac{2}{7} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{5} d u - \int 3 u^{4} d u - \int u^{2} d u + \int \frac{2}{7} d u

\int u^{5} d u = \frac{u ^{6}}{6}

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int \frac{2}{7} d u = \frac{2}{7} u

= \frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u

Add a constant to the solution

= \frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u + C

- \frac{3}{x}

\frac{d}{d x} (2 x sin (x) + cos (x) x^{2})

x \to \infty lim (3 \cdot x)

\int \frac{17}{x ln ( 3 x )} d x

\frac{d}{d x} (2 - x^{2})

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