What is the integral of 10(1+x/(60))^2 ?

The integral of 10(1+x/(60))^2 is 10(x+(x^2)/(60)+(x^3)/(10800))+C

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integral of 10(1+x/(60))^2

\int 10 (1 + \frac{x}{60})^{2} d x

10 (x + \frac{x ^{2}}{60} + \frac{x ^{3}}{10800}) + C

Solution steps

\int 10 (1 + \frac{x}{60})^{2} d x

Take the constant out:

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

= 10 \cdot \int (1 + \frac{x}{60})^{2} d x

Expand

(1 + \frac{x}{60})^{2} : 1 + \frac{x}{30} + \frac{x ^{2}}{3600}

= 10 \cdot \int 1 + \frac{x}{30} + \frac{x ^{2}}{3600} 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

= 10 (\int 1 d x + \int \frac{x}{30} d x + \int \frac{x ^{2}}{3600} d x)

\int 1 d x = x

\int \frac{x}{30} d x = \frac{x ^{2}}{60}

\int \frac{x ^{2}}{3600} d x = \frac{x ^{3}}{10800}

= 10 (x + \frac{x ^{2}}{60} + \frac{x ^{3}}{10800})

Add a constant to the solution

= 10 (x + \frac{x ^{2}}{60} + \frac{x ^{3}}{10800}) + C

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What is the integral of 10(1+x/(60))^2 ?
The integral of 10(1+x/(60))^2 is 10(x+(x^2)/(60)+(x^3)/(10800))+C