What is the integral from 0 to x of ln(z+2) ?

The integral from 0 to x of ln(z+2) is xln(x+2)+2ln(x+2)-x-2ln(2)

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integral from 0 to x of ln(z+2)

\int_{0}^{x} ln (z + 2) d z

x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

Solution steps

\int_{0}^{x} ln (z + 2) d z

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Compute the boundaries:

x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

= x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

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What is the integral from 0 to x of ln(z+2) ?
The integral from 0 to x of ln(z+2) is xln(x+2)+2ln(x+2)-x-2ln(2)