What is the integral from 0 to 5 of ln(1+x) ?

The integral from 0 to 5 of ln(1+x) is 6ln(6)-5

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integral from 0 to 5 of ln(1+x)

\int_{0}^{5} ln (1 + x) d x

6 ln (6) - 5

Decimal Notation

5.75055 \dots

Solution steps

\int_{0}^{5} ln (1 + x) d x

Apply u-substitution

: \int_{1}^{6} ln (u) d u

= \int_{1}^{6} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{1}^{6}

= [u ln (u) - \int 1 d u]_{1}^{6}

\int 1 d u = u

= [u ln (u) - u]_{1}^{6}

Compute the boundaries:

6 ln (6) - 5

= 6 ln (6) - 5

\frac{d}{d x} (2 x^{3} - 7 x^{2} + 3 x - 2)

\frac{x - 10}{x - 10}

\frac{d}{d x} ((7 + 2 x) e^{- 3 x})

t^{2} y^{^{''}} + 2 t y^{^{'}} - 1 = 0

\frac{d}{d x} (10 x (x^{2} + 5)^{4})

What is the integral from 0 to 5 of ln(1+x) ?
The integral from 0 to 5 of ln(1+x) is 6ln(6)-5