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

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

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

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

- 1 + \frac{1}{t}

Solution steps

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

Apply the Power Rule

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

Compute the boundaries:

- 1 + \frac{1}{t}

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

(t^{2} + 5)^{2} - 7 (t^{2} + 5)

2 x \frac{d y}{d x} - 8 y = x^{- 8}

\frac{d y}{d x} = - \frac{x}{y}, y (0) = 1

\frac{\partial}{\partial x} (e^{x y} + z)

\int_{- 10}^{10} x^{2} d x

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