What is the integral from 0 to 8 of 1/(sqrt(x+1)) ?

The integral from 0 to 8 of 1/(sqrt(x+1)) is 4

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integral from 0 to 8 of 1/(sqrt(x+1))

\int_{0}^{8} \frac{1}{x + 1} d x

4

Solution steps

\int_{0}^{8} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{1}^{3} 2 d u

= \int_{1}^{3} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{1}^{3}

Compute the boundaries:

4

= 4

\int_{1}^{10} \frac{1}{6 x + 3} d x

\int_{1}^{e} 9 t^{2} ln (t) d t

\int_{0}^{49} x d x

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

\int_{0}^{1} 2 x (5 - x^{2}) d x

What is the integral from 0 to 8 of 1/(sqrt(x+1)) ?
The integral from 0 to 8 of 1/(sqrt(x+1)) is 4