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

The integral from 0 to 4 of ln(1+sqrt(x)) is 3ln(3)

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integral from 0 to 4 of ln(1+sqrt(x))

\int_{0}^{4} ln (1 + x) d x

3 ln (3)

Decimal Notation

3.29583 \dots

Solution steps

\int_{0}^{4} ln (1 + x) d x

Apply u-substitution

: \int_{1}^{3} ln (u) \cdot 2 (u - 1) d u

= \int_{1}^{3} 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}^{3} ln (u) (u - 1) d u

Apply Integration By Parts

: [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{1}^{3}

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{1}^{3}

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3}

Simplify

2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3} : 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

= 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

Compute the boundaries:

\frac{3}{2} ln (3)

= 2 \cdot \frac{3}{2} ln (3)

Simplify

= 3 ln (3)

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\int_{0}^{x} t^{k} d t

What is the integral from 0 to 4 of ln(1+sqrt(x)) ?
The integral from 0 to 4 of ln(1+sqrt(x)) is 3ln(3)