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

The integral from 0 to 1 of ln(2x+1) is 3/2 ln(3)-1

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

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

\frac{3}{2} ln (3) - 1

Decimal Notation

0.64791 \dots

Solution steps

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

Apply u-substitution

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

= \int_{1}^{3} 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}^{3} ln (u) d u

Apply Integration By Parts

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

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

\int 1 d u = u

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

Compute the boundaries:

3 ln (3) - 2

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

Simplify

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

f (x) = - 4 x^{2} - 3 + 2, at x = - 1

\int_{0}^{1} \frac{x + 1}{x ^{2} + 1} d x

\int \frac{e ^{\frac{1}{x ^{8}}}}{x ^{9}} d x

x \to - 3 lim (\frac{x - 3}{9 x})

\frac{\partial}{\partial x} (x^{2} + 5 x z + 3 y z^{2})

What is the integral from 0 to 1 of ln(2x+1) ?
The integral from 0 to 1 of ln(2x+1) is 3/2 ln(3)-1