Solve \int_{0}^{\frac{1}{2}}\left(-1+\frac{1}{1+x}+\f

Evaluate

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

Evaluate the integral

\int (- 1 + \frac{1}{1 + x} + \frac{1}{1 - x}) d x

Use the property of integral \int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int - 1 d x + \int \frac{1}{1 + x} d x + \int \frac{1}{1 - x} d x

Use the property of integral \int k d x = k x

- x + \int \frac{1}{1 + x} d x + \int \frac{1}{1 - x} d x

Use the property of integral \int \frac{1}{a x + b} d x = \frac{1}{a} ln ∣ a x + b ∣

- x + ln (∣ x + 1 ∣) + \int \frac{1}{1 - x} d x

Evaluate the integral

More Steps

- x + ln (∣ x + 1 ∣) - ln (∣ x - 1 ∣)

Use lo g_{a} x - lo g_{a} y = lo g_{a} \frac{x}{y} to transform the expression

- x + ln (\frac{∣ x + 1 ∣}{∣ x - 1 ∣})

Return the limits

(- x + ln (\frac{∣ x + 1 ∣}{∣ x - 1 ∣}))_{0}^{\frac{1}{2}}

Solution

More Steps

Substitute the values into formula

- \frac{1}{2} + ln (\frac{\frac{1}{2} + 1}{\frac{1}{2} - 1}) - (- 0 + ln (\frac{∣ 0 + 1 ∣}{∣ 0 - 1 ∣}))

Evaluate

- \frac{1}{2} + ln (\frac{\frac{1}{2} + 1}{\frac{1}{2} - 1}) - (0 + ln (\frac{∣ 0 + 1 ∣}{∣ 0 - 1 ∣}))

Removing 0 doesn't change the value,so remove it from the expression

- \frac{1}{2} + ln (\frac{\frac{1}{2} + 1}{\frac{1}{2} - 1}) - (0 + ln (\frac{∣ 1 ∣}{∣ 0 - 1 ∣}))

Removing 0 doesn't change the value,so remove it from the expression

- \frac{1}{2} + ln (\frac{\frac{1}{2} + 1}{\frac{1}{2} - 1}) - (0 + ln (\frac{∣ 1 ∣}{∣ - 1 ∣}))

Add the numbers

More Steps

- \frac{1}{2} + ln (\frac{\frac{3}{2}}{\frac{1}{2} - 1}) - (0 + ln (\frac{∣ 1 ∣}{∣ - 1 ∣}))

Subtract the numbers

More Steps

- \frac{1}{2} + ln (\frac{\frac{3}{2}}{- \frac{1}{2}}) - (0 + ln (\frac{∣ 1 ∣}{∣ - 1 ∣}))

When the expression in absolute value bars is not negative, remove the bars

- \frac{1}{2} + ln (\frac{\frac{3}{2}}{- \frac{1}{2}}) - (0 + ln (\frac{1}{∣ - 1 ∣}))

Since - 1 <0,the absolute value of - 1 is 1

- \frac{1}{2} + ln (\frac{\frac{3}{2}}{- \frac{1}{2}}) - (0 + ln (\frac{1}{1}))

When the expression in absolute value bars is not negative, remove the bars

- \frac{1}{2} + ln (\frac{\frac{3}{2}}{- \frac{1}{2}}) - (0 + ln (\frac{1}{1}))

Since - \frac{1}{2} <0,the absolute value of - \frac{1}{2} is \frac{1}{2}

- \frac{1}{2} + ln (\frac{\frac{3}{2}}{\frac{1}{2}}) - (0 + ln (\frac{1}{1}))

Divide the terms

- \frac{1}{2} + ln (\frac{\frac{3}{2}}{\frac{1}{2}}) - (0 + ln (1))

Divide the terms

More Steps

- \frac{1}{2} + ln (3) - (0 + ln (1))

Evaluate the logarithm

- \frac{1}{2} + ln (3) - (0 + 0)

Removing 0 doesn't change the value,so remove it from the expression

- \frac{1}{2} + ln (3) - 0

Removing 0 doesn't change the value,so remove it from the expression

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

Simplify

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

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

Alternative Form

\approx 0.598612