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

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

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

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

2 ln (2)

Decimal

1.38629 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{1}{u} d u

= \int_{1}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= 2 ln (2)

x \to 0 + lim (\frac{x ^{2}}{7} - \frac{2}{x})

x \to 0 lim ((1 + 3 x)^{\frac{2}{3}})

\int (x^{2} + 2 x + 3) cos (2 x) d x

\int_{- \infty}^{0} 5^{x} d x

y = 9 - 5 x^{2}

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