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

The integral from 0 to 1 of (1+x)/(1+2x) is (ln(3)+2)/4

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

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

\frac{ln ( 3 ) + 2}{4}

Decimal Notation

0.77465 \dots

Solution steps

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

Expand

\frac{1 + x}{1 + 2 x} : \frac{1}{1 + 2 x} + \frac{x}{1 + 2 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} \frac{1}{1 + 2 x} d x = \frac{1}{2} ln (3)

\int_{0}^{1} \frac{x}{1 + 2 x} d x = \frac{1}{4} (2 - ln (3))

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

Simplify

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

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

\frac{e ^{x}}{1 + x}

\frac{d y}{d x} + a y + b = 0

x \to 3 - lim (\frac{x + 3}{2})

\int_{0}^{x^{4}} cos (t^{2}) d t

(2, 5), (7, 3)

What is the integral from 0 to 1 of (1+x)/(1+2x) ?
The integral from 0 to 1 of (1+x)/(1+2x) is (ln(3)+2)/4