What is the integral from-7 to-3 of 1/x ?

The integral from-7 to-3 of 1/x is ln(3)-ln(7)

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integral from-7 to-3 of 1/x

\int_{- 7}^{- 3} \frac{1}{x} d x

ln (3) - ln (7)

Decimal Notation

- 0.84729 \dots

Solution steps

\int_{- 7}^{- 3} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{- 7}^{- 3}

Compute the boundaries:

ln (3) - ln (7)

= ln (3) - ln (7)

\int - e^{x} cos (y) d y

\frac{\partial}{\partial x} (4 x + 6 y + 2 z)

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} ln (t)

\int \frac{3}{8} x^{2} d x

\frac{d y}{d x} = \frac{x + 3 y}{y - 3 x}

What is the integral from-7 to-3 of 1/x ?
The integral from-7 to-3 of 1/x is ln(3)-ln(7)