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

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

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

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

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

Decimal Notation

0.23661 \dots

Solution steps

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

Take the partial fraction of

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

= \int_{1}^{e} \frac{1}{x} - \frac{2}{2 x + 1} d 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_{1}^{e} \frac{1}{x} d x - \int_{1}^{e} \frac{2}{2 x + 1} d x

\int_{1}^{e} \frac{1}{x} d x = 1

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

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

Simplify

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

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