What is the integral of (x^3+1)^2(3x) ?

The integral of (x^3+1)^2(3x) is 3((x^8)/8+(2x^5)/5+(x^2)/2)+C

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integral of (x^3+1)^2(3x)

\int (x^{3} + 1)^{2} (3 x) d x

3 (\frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2}) + C

Solution steps

\int (x^{3} + 1)^{2} \cdot 3 x d x

Take the constant out:

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

= 3 \cdot \int (x^{3} + 1)^{2} x d x

Expand

(x^{3} + 1)^{2} x : x^{7} + 2 x^{4} + x

= 3 \cdot \int x^{7} + 2 x^{4} + x 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

= 3 (\int x^{7} d x + \int 2 x^{4} d x + \int x d x)

\int x^{7} d x = \frac{x ^{8}}{8}

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

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

= 3 (\frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2})

Add a constant to the solution

= 3 (\frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2}) + C

y = \frac{x}{x + 9}, (1, 0.1)

- 2 x^{\frac{2}{3}} - 4, at x = 8

(2 x^{2} - 4)^{3}

x \to 0 - lim (\frac{cot ( π x ) sin ( x )}{2 sec ( x )})

n = 0 \sum \infty 2 n - 1

What is the integral of (x^3+1)^2(3x) ?
The integral of (x^3+1)^2(3x) is 3((x^8)/8+(2x^5)/5+(x^2)/2)+C