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

The integral from 1 to 2 of (ax^3+6/(x^2)) is (15a)/4+3

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

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

\frac{15 a}{4} + 3

Solution steps

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

\int_{1}^{2} a x^{3} d x = \frac{15 a}{4}

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

= \frac{15 a}{4} + 3

What is the integral from 1 to 2 of (ax^3+6/(x^2)) ?
The integral from 1 to 2 of (ax^3+6/(x^2)) is (15a)/4+3