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

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

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

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

4

Solution steps

\int_{0}^{8} \frac{1}{1 + x} 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 2 x y - 9 x^{2} d x

\int_{2}^{4} 10 + 4 x - 3 x^{3} d x

2 x - 2 x^{2}

0.9 x^{3} 3^{x}

\frac{\partial}{\partial x} (\frac{4 x}{3 y ^{2} + 1})

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