What is the integral from 0 to 3 of 1/(1+x) ?

The integral from 0 to 3 of 1/(1+x) is 2ln(2)

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integral from 0 to 3 of 1/(1+x)

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

2 ln (2)

Solution steps

\int_{0}^{3} \frac{1}{1 + x} 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)

\int_{x^{2}}^{x} (x^{2} + y^{2}) d y

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

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

\int_{0}^{2} z d z

\int_{0}^{1} x (x^{2} + 6) d x

What is the integral from 0 to 3 of 1/(1+x) ?
The integral from 0 to 3 of 1/(1+x) is 2ln(2)