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

The integral from 3 to 6 of (1/(x^2)-1) is -17/6

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

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

- \frac{17}{6}

Decimal

- 2.83333 \dots

Solution steps

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

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

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

= \frac{1}{6} - 3

Simplify

\frac{1}{6} - 3 : - \frac{17}{6}

= - \frac{17}{6}

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

\int_{- 2}^{2} (x + 17)^{2} d x

\int_{1}^{6} 8 d x

\int_{2}^{4} 3 x^{4} d x

\int_{1}^{6} 6 d x

What is the integral from 3 to 6 of (1/(x^2)-1) ?
The integral from 3 to 6 of (1/(x^2)-1) is -17/6