What is the integral from 0 to 2x of ln(1+t) ?

The integral from 0 to 2x of ln(1+t) is ln(2x+1)+2xln(2x+1)-2x

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integral from 0 to 2x of ln(1+t)

\int_{0}^{2 x} ln (1 + t) d t

ln (2 x + 1) + 2 x ln (2 x + 1) - 2 x

Solution steps

\int_{0}^{2 x} ln (1 + t) d t

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Compute the boundaries:

ln (2 x + 1) + 2 x ln (2 x + 1) - 2 x

= ln (2 x + 1) + 2 x ln (2 x + 1) - 2 x

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What is the integral from 0 to 2x of ln(1+t) ?
The integral from 0 to 2x of ln(1+t) is ln(2x+1)+2xln(2x+1)-2x