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

The integral from 1 to 2e of ln(2x+1) is 2eln(4e+1)+1/2 ln(4e+1)+1-2e-3/2 ln(3)

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

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

2 e ln (4 e + 1) + \frac{1}{2} ln (4 e + 1) + 1 - 2 e - \frac{3}{2} ln (3)

Decimal

8.60422 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{4 e + 1} ln (u) \frac{1}{2} d u

= \int_{3}^{4 e + 1} 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_{3}^{4 e + 1} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{3}^{4 e + 1}

= \frac{1}{2} [u ln (u) - \int 1 d u]_{3}^{4 e + 1}

\int 1 d u = u

= \frac{1}{2} [u ln (u) - u]_{3}^{4 e + 1}

Compute the boundaries:

4 e ln (4 e + 1) + ln (4 e + 1) + 2 - 4 e - 3 ln (3)

= \frac{1}{2} (4 e ln (4 e + 1) + ln (4 e + 1) + 2 - 4 e - 3 ln (3))

Simplify

= 2 e ln (4 e + 1) + \frac{1}{2} ln (4 e + 1) + 1 - 2 e - \frac{3}{2} ln (3)

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What is the integral from 1 to 2e of ln(2x+1) ?
The integral from 1 to 2e of ln(2x+1) is 2eln(4e+1)+1/2 ln(4e+1)+1-2e-3/2 ln(3)