What is the integral of (x^4+x+9) ?

The integral of (x^4+x+9) is (x^5)/5+(x^2)/2+9x+C

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integral of (x^4+x+9)

\int (x^{4} + x + 9) d x

\frac{x ^{5}}{5} + \frac{x ^{2}}{2} + 9 x + C

Solution steps

\int x^{4} + x + 9 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 x^{4} d x + \int x d x + \int 9 d x

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

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

\int 9 d x = 9 x

= \frac{x ^{5}}{5} + \frac{x ^{2}}{2} + 9 x

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{x ^{2}}{2} + 9 x + C

x \to - 1 lim (3 x - 5)

ln (\frac{x ^{2} - 1}{x ^{2} + 1})

\frac{s ^{2} + 1}{s ^{2} ( s + 2 )}

\int 2 x (3 x - 10)^{4} d x

2 x - 2 y

What is the integral of (x^4+x+9) ?
The integral of (x^4+x+9) is (x^5)/5+(x^2)/2+9x+C