What is the integral from 1 to 2 of (3x^3+6/(x^2)) ?

The integral from 1 to 2 of (3x^3+6/(x^2)) is 57/4

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integral from 1 to 2 of (3x^3+6/(x^2))

\int_{1}^{2} (3 x^{3} + \frac{6}{x ^{2}}) d x

\frac{57}{4}

Solution steps

\int_{1}^{2} 3 x^{3} + \frac{6}{x ^{2}} 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

= \int_{1}^{2} 3 x^{3} d x + \int_{1}^{2} \frac{6}{x ^{2}} d x

\int_{1}^{2} 3 x^{3} d x = \frac{45}{4}

\int_{1}^{2} \frac{6}{x ^{2}} d x = 3

= \frac{45}{4} + 3

Simplify

\frac{45}{4} + 3 : \frac{57}{4}

= \frac{57}{4}

\int_{- 2}^{2} x^{5} d x

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

\int_{0}^{5} 6

\int_{1}^{2} (4 x^{3} - 6) d x

\int_{\frac{1}{3}}^{3} 5 - \frac{5}{x ^{2}} d x

What is the integral from 1 to 2 of (3x^3+6/(x^2)) ?
The integral from 1 to 2 of (3x^3+6/(x^2)) is 57/4