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

The integral from-3 to-2 of 1/(1+x) is -ln(2)

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

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

- ln (2)

Decimal Notation

- 0.69314 \dots

Solution steps

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

Apply u-substitution

: \int_{- 2}^{- 1} \frac{1}{u} d u

= \int_{- 2}^{- 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 2}^{- 1}

Compute the boundaries:

- ln (2)

= - ln (2)

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 3)^{2} d x

\int_{0}^{1} u d u

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

\int_{50}^{100} \frac{x}{17 + 6 x ^{2}} d x

\int_{0}^{1} x^{3} x^{2} + 1 d x

What is the integral from-3 to-2 of 1/(1+x) ?
The integral from-3 to-2 of 1/(1+x) is -ln(2)