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

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

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integral of 5x(x^2+1)

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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