What is the integral of u^5-6u^4-u^2+4/3 ?

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

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integral of u^5-6u^4-u^2+4/3

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

\frac{u ^{6}}{6} - \frac{6 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{4}{3} u + C

Solution steps

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

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

\int 6 u^{4} d u = \frac{6 u ^{5}}{5}

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

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

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

Add a constant to the solution

= \frac{u ^{6}}{6} - \frac{6 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{4}{3} u + C

f (x) = ln (x) - x

x \to - \infty lim (0)

\int sec^{2} (x) - 1 d x

y = \frac{x}{x + 1}

x \to - 9 + lim (\frac{x + 8}{x + 9})

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