What is the integral of 5x(x^2+1)^3 ?

The integral of 5x(x^2+1)^3 is 5((x^8)/8+(x^6)/2+(3x^4)/4+(x^2)/2)+C

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integral of 5x(x^2+1)^3

\int 5 x (x^{2} + 1)^{3} d x

5 (\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2}) + C

Solution steps

\int 5 x (x^{2} + 1)^{3} d x

Take the constant out:

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

= 5 \cdot \int x (x^{2} + 1)^{3} d x

Expand

x (x^{2} + 1)^{3} : x^{7} + 3 x^{5} + 3 x^{3} + x

= 5 \cdot \int x^{7} + 3 x^{5} + 3 x^{3} + 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

= 5 (\int x^{7} d x + \int 3 x^{5} d x + \int 3 x^{3} d x + \int x d x)

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

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

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

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

= 5 (\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2})

Add a constant to the solution

= 5 (\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2}) + C

\int_{- 3}^{2} 6 - x - x^{2} d x

\frac{d}{d x} (2 x^{4} + 7 x + 5)

x \to - \infty lim (\frac{7}{x})

\int x (x + 2) d x

(ln (5 x))^{2}

What is the integral of 5x(x^2+1)^3 ?
The integral of 5x(x^2+1)^3 is 5((x^8)/8+(x^6)/2+(3x^4)/4+(x^2)/2)+C