What is the integral from 1 to 8 of 3x^2+1/(12x^2) ?

The integral from 1 to 8 of 3x^2+1/(12x^2) is 49063/96

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integral from 1 to 8 of 3x^2+1/(12x^2)

\int_{1}^{8} 3 x^{2} + \frac{1}{12 x ^{2}} d x

\frac{49063}{96}

Solution steps

\int_{1}^{8} 3 x^{2} + \frac{1}{12 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}^{8} 3 x^{2} d x + \int_{1}^{8} \frac{1}{12 x ^{2}} d x

\int_{1}^{8} 3 x^{2} d x = 511

\int_{1}^{8} \frac{1}{12 x ^{2}} d x = \frac{7}{96}

= 511 + \frac{7}{96}

Simplify

511 + \frac{7}{96} : \frac{49063}{96}

= \frac{49063}{96}

\int \frac{x ^{2}}{( x - 1 ) ^{4}} d x

\frac{x ^{3} - 4}{x ^{2} + 2}

f (t) = \frac{1}{t}

x^{2} + 5 x

x ln (2 x) y^{^{'}} - y = 0

What is the integral from 1 to 8 of 3x^2+1/(12x^2) ?
The integral from 1 to 8 of 3x^2+1/(12x^2) is 49063/96