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

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

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

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

2

Solution steps

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

Apply u-substitution

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

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

Integral of a constant:

\int a d x = a x

= [2 u]_{1}^{2}

Compute the boundaries:

2

= 2

\int \frac{z ^{2}}{1 ^{3}} + z^{3}

\int 20 x^{3} - 12 x^{2} + 6 x d x

n = 1 \sum \infty \frac{7}{n ^{2} + 9}

x \to 0 + lim (\frac{e ^{x}}{e ^{x} - 1})

\frac{\partial}{\partial y} (e^{y^{2}} - e^{y})

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