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

The integral from 1 to 8 of 1/(x^2) is 7/8

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

\int_{1}^{8} \frac{1}{x ^{2}} d x

\frac{7}{8}

Decimal Notation

0.875

Solution steps

\int_{1}^{8} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{8}

= [- \frac{1}{x}]_{1}^{8}

Compute the boundaries:

\frac{7}{8}

= \frac{7}{8}

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

\int_{2}^{3} x^{2} - 5 x + 6 d x

\int_{0}^{1} \frac{1}{( 1 - x ) ^{2}} d x

\int_{0}^{1} (x - 1) ln (x) d x

\int_{0}^{\frac{3 π}{2}} x sin (x) d x

What is the integral from 1 to 8 of 1/(x^2) ?
The integral from 1 to 8 of 1/(x^2) is 7/8