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

The integral from 1 to t of 1/(x^2) is -1/t+1

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

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

- \frac{1}{t} + 1

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{t} + 1

= - \frac{1}{t} + 1

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\int_{1}^{5} \frac{e ^{\frac{5}{x}}}{x ^{2}} d x

\int_{x^{5}}^{x^{4}} (2 t - 1)^{3} d t

\int_{0}^{\frac{π}{2}} - sin (x) d x

What is the integral from 1 to t of 1/(x^2) ?
The integral from 1 to t of 1/(x^2) is -1/t+1