What is the integral from 3 to sqrt(x of)(t^2+1) ?

The integral from 3 to sqrt(x of)(t^2+1) is sqrt(x)+(x^{3/2})/3-12

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integral from 3 to sqrt(x of)(t^2+1)

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

x + \frac{x ^{\frac{3}{2}}}{3} - 12

Solution steps

\int_{3}^{x} t^{2} + 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{3}^{x} t^{2} d t + \int_{3}^{x} 1 d t

\int_{3}^{x} t^{2} d t = \frac{x ^{\frac{3}{2}}}{3} - 9

\int_{3}^{x} 1 d t = x - 3

= \frac{x ^{\frac{3}{2}}}{3} - 9 + x - 3

Simplify

= x + \frac{x ^{\frac{3}{2}}}{3} - 12

What is the integral from 3 to sqrt(x of)(t^2+1) ?
The integral from 3 to sqrt(x of)(t^2+1) is sqrt(x)+(x^{3/2})/3-12