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

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

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

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

\frac{1}{2} ln (2 x + 1) + x ln (2 x + 1) - x

Solution steps

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

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

\int 1 d u = u

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

Compute the boundaries:

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

= \frac{1}{2} (ln (2 x + 1) + 2 x ln (2 x + 1) - 2 x)

Simplify

= \frac{1}{2} ln (2 x + 1) + x ln (2 x + 1) - x

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