What is the integral from 1 to 2 of ln(x+1) ?

The integral from 1 to 2 of ln(x+1) is 3ln(3)-1-2ln(2)

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integral from 1 to 2 of ln(x+1)

\int_{1}^{2} ln (x + 1) d x

3 ln (3) - 1 - 2 ln (2)

Decimal Notation

0.90954 \dots

Solution steps

\int_{1}^{2} ln (x + 1) d x

Apply u-substitution

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

= \int_{2}^{3} ln (u) d u

Apply Integration By Parts

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

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

\int 1 d u = u

= [u ln (u) - u]_{2}^{3}

Compute the boundaries:

3 ln (3) - 1 - 2 ln (2)

= 3 ln (3) - 1 - 2 ln (2)

\int (x^{7} + 1) e^{x^{8} + 8 x} d x

\int \frac{( e ^{- x} )}{x} d x

t \frac{d y}{d t} = y + 4 t^{2} sin (3 t)

\frac{\partial}{\partial y} (\frac{x + y ^{2}}{z})

What is the integral from 1 to 2 of ln(x+1) ?
The integral from 1 to 2 of ln(x+1) is 3ln(3)-1-2ln(2)