What is the integral of 9x^3(x^3+2x^2+x+8) ?

The integral of 9x^3(x^3+2x^2+x+8) is 9((x^7)/7+(x^6)/3+(x^5)/5+2x^4)+C

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integral of 9x^3(x^3+2x^2+x+8)

\int 9 x^{3} (x^{3} + 2 x^{2} + x + 8) d x

9 (\frac{x ^{7}}{7} + \frac{x ^{6}}{3} + \frac{x ^{5}}{5} + 2 x^{4}) + C

Solution steps

\int 9 x^{3} (x^{3} + 2 x^{2} + x + 8) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int x^{3} (x^{3} + 2 x^{2} + x + 8) d x

Expand

x^{3} (x^{3} + 2 x^{2} + x + 8) : x^{6} + 2 x^{5} + x^{4} + 8 x^{3}

= 9 \cdot \int x^{6} + 2 x^{5} + x^{4} + 8 x^{3} 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

= 9 (\int x^{6} d x + \int 2 x^{5} d x + \int x^{4} d x + \int 8 x^{3} d x)

\int x^{6} d x = \frac{x ^{7}}{7}

\int 2 x^{5} d x = \frac{x ^{6}}{3}

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

\int 8 x^{3} d x = 2 x^{4}

= 9 (\frac{x ^{7}}{7} + \frac{x ^{6}}{3} + \frac{x ^{5}}{5} + 2 x^{4})

Add a constant to the solution

= 9 (\frac{x ^{7}}{7} + \frac{x ^{6}}{3} + \frac{x ^{5}}{5} + 2 x^{4}) + C

x \to 4 - lim (c x^{2} + 7 x)

\int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x

f (x) = \frac{2}{3 x ^{2}}

n = 1 \sum \infty 2^{- 2 n}

What is the integral of 9x^3(x^3+2x^2+x+8) ?
The integral of 9x^3(x^3+2x^2+x+8) is 9((x^7)/7+(x^6)/3+(x^5)/5+2x^4)+C