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

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

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integral from 0 to pi of 1/(2x+pi)

\int_{0}^{π} \frac{1}{2 x + π} d x

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

Decimal Notation

0.54930 \dots

Solution steps

\int_{0}^{π} \frac{1}{2 x + π} d x

Apply u-substitution

: \int_{π}^{3 π} \frac{1}{2 u} d u

= \int_{π}^{3 π} \frac{1}{2 u} 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_{π}^{3 π} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{π}^{3 π}

Compute the boundaries:

ln (3 π) - ln (π)

= \frac{1}{2} (ln (3 π) - ln (π))

Simplify

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

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

\int_{- 1}^{2} x^{3} + 8 d x

\int_{- π}^{π} ∣ t ∣ d t

\int_{0}^{y} x^{2} + y^{2} d x

\int_{0}^{x} 4 + t^{6} d x

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