What is the integral from 0 to 0.64 of ln(1+sqrt(x)) ?

The integral from 0 to 0.64 of ln(1+sqrt(x)) is 0.26839…

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integral from 0 to 0.64 of ln(1+sqrt(x))

\int_{0}^{0.64} ln (1 + x) d x

0.26839 \dots

Solution steps

\int_{0}^{0.64} ln (1 + x) d x

Apply u-substitution

= \int_{1}^{1.8} ln (u) \cdot 2 (u - 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{1.8} ln (u) (u - 1) d u

Apply Integration By Parts

= 2 [ln (u) (0.5 u^{2} - u) - \int 0.5 u - 1 d u]_{1}^{1.8}

\int 0.5 u - 1 d u = 0.25 u^{2} - u

= 2 [ln (u) (0.5 u^{2} - u) - (0.25 u^{2} - u)]_{1}^{1.8}

Expand

ln (u) (0.5 u^{2} - u) - (0.25 u^{2} - u) : 0.5 u^{2} ln (u) - u ln (u) - 0.25 u^{2} + u

= 2 [0.5 u^{2} ln (u) - u ln (u) - 0.25 u^{2} + u]_{1}^{1.8}

Compute the boundaries:

0.13419 \dots

= 2 \cdot 0.13419 \dots

Simplify

= 0.26839 \dots

What is the integral from 0 to 0.64 of ln(1+sqrt(x)) ?
The integral from 0 to 0.64 of ln(1+sqrt(x)) is 0.26839…