integral from -1/4 to 1/4 of 1/(1-x^2)

1) Let's integrate:

\int_{- \frac{1}{4}}^{\frac{1}{4}} \frac{1}{1 - x^{2}} d x

2) Let's factor in

ℝ

1 - x^{2}

For

1 - x^{2}

, we recognize a difference of two squares:

1^{2} - x^{2}

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = 1

and

b = x

Factors found:

(1 - x) \cdot (x + 1)

3) You can decompose into partial fractions:

\frac{1}{(1 - x) \cdot (x + 1)} = \frac{A}{1 - x} + \frac{B}{x + 1}

For the simple factor:

1 - x

and its root:

1

, the unknown A can be easily computed by multiplying by the denominator:

1 - x^{2}

and evaluating at the root because the other terms cancel out:

4) Let's solve in

ℝ

(1 + 1) \cdot A = 1

ⓘ Parentheses are computed first:

1 + 1

2

2 A = 1

ⓘ Both sides are divided by

2

Solution obtained:

A = \frac{1}{2}

For the simple factor:

x + 1

and its root:

-1

, compute B by multiplying by the denominator:

1 - x^{2}

and evaluating at the root (the other terms cancel out):

5) Let's solve in

ℝ

(1 + 1) \cdot B = 1

ⓘ Parentheses are computed first:

1 + 1

2

2 B = 1

ⓘ Both sides are divided by

2

Solution obtained:

B = \frac{1}{2}

6) Finally, substitute the unknowns back into the initial partial fraction decomposition:

\frac{1}{(1 - x) \cdot (x + 1)} = \frac{1}{2 \cdot (1 - x)} + \frac{1}{2 \cdot (x + 1)}

7) Let's integrate:

\int (\frac{1}{2 \cdot (1 - x)} + \frac{1}{2 \cdot (x + 1)}) d x

ⅰ

\int (\frac{1}{2 \cdot (1 - x)} + \frac{1}{2 \cdot (x + 1)}) d x

\int \frac{1}{2 \cdot (1 - x)} d x + \int \frac{1}{2 \cdot (x + 1)} d x

\int \frac{1}{2 \cdot (1 - x)} d x + \int \frac{1}{2 \cdot (x + 1)} d x

Let

u = 2 - 2 x

. Then

du = - 2 dx

, so

dx = - \frac{du}{2}

\int \frac{1}{2 \cdot (1 - x)} d x = - \frac{1}{2} \cdot \int \frac{1}{u} d u

We use

\int \frac{1}{u} d u = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

2 - 2 x \neq 0

Substitute

u

2 - 2 x

back into the antiderivative

Let

u = 2 x + 2

. Then

du = 2 dx

, so

dx = \frac{du}{2}

\int \frac{1}{2 \cdot (x + 1)} d x = \frac{1}{2} \cdot \int \frac{1}{u} d u

We use

\int \frac{1}{u} d u = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

2 x + 2 \neq 0

Substitute

u

2 x + 2

back into the antiderivative

ⓘ The derivative of

\frac{1}{2} \cdot \ln | 2 x + 2 |

\frac{1}{2 \cdot (x + 1)}

, so

\frac{1}{2} \cdot \ln | 2 x + 2 |

is an antiderivative of

\frac{1}{2 \cdot (x + 1)}

\int \frac{1}{2 \cdot (x + 1)} d x

\frac{1}{2} \cdot \ln | 2 x + 2 |

- \frac{1}{2} \cdot \ln | 2 - 2 x | + \frac{1}{2} \cdot \ln | 2 x + 2 |

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot \ln | 2 - 2 x |

\frac{1 \cdot \ln | 2 - 2 x |}{2}

\frac{\ln | 2 - 2 x |}{2}

\frac{1}{2} \cdot \ln | 2 x + 2 |

\frac{1 \cdot \ln | 2 x + 2 |}{2}

\frac{\ln | 2 x + 2 |}{2}

■For strictly positive arguments, a sum of logarithms is the logarithm of the product, a difference that of the quotient, ln(a) + ln(b) = ln(a*b):

- \frac{\ln | 2 - 2 x |}{2} + \frac{\ln | 2 x + 2 |}{2}

\ln (\frac{\sqrt{| 2 x + 2 |}}{\sqrt{| 2 - 2 x |}})

ⅰ

\int \frac{1}{1 - x^{2}} d x

\ln (\frac{\sqrt{| 2 x + 2 |}}{\sqrt{| 2 - 2 x |}})

▷

\ln (\frac{\sqrt{| 2 x + 2 |}}{\sqrt{| 2 - 2 x |}})

8) Let's evaluate over the interval

[- \frac{1}{4}, \frac{1}{4}]

\ln (\frac{\sqrt{| 2 \cdot \frac{1}{4} + 2 |}}{\sqrt{| 2 - 2 \cdot \frac{1}{4} |}}) - \ln (\frac{\sqrt{| 2 \cdot (- \frac{1}{4}) + 2 |}}{\sqrt{| 2 - 2 \cdot (- \frac{1}{4}) |}})

ⓘ Multiplying a fraction by a number multiplies its numerator:

2 \cdot \frac{1}{4}

\frac{2 \cdot 1}{4}

\frac{2}{4}

■We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{4}

\frac{1 \cdot 2}{2 \cdot 2}

\frac{1}{2}

- \frac{2}{4}

- \frac{1 \cdot 2}{2 \cdot 2}

- \frac{1}{2}

■The minus sign is placed in front of the product:

2 \cdot (- \frac{1}{4})

- 2 \cdot \frac{1}{4}

■Two minus signs cancel out, the product of two negative numbers is positive:

- 2 \cdot (- \frac{1}{4})

2 \cdot \frac{1}{4}

\ln (\frac{\sqrt{| \frac{1}{2} + 2 |}}{\sqrt{| 2 - \frac{1}{2} |}}) - \ln (\frac{\sqrt{| - 2 \cdot \frac{1}{4} + 2 |}}{\sqrt{| 2 + 2 \cdot \frac{1}{4} |}})

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1}{2} + 2

\frac{1 + 2 \cdot 2}{2}

2 - \frac{1}{2}

\frac{2 \cdot 2 - 1}{2}

■Multiplying a fraction by a number multiplies its numerator:

2 \cdot \frac{1}{4}

\frac{2 \cdot 1}{4}

\frac{2}{4}

■We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{2}{4}

- \frac{1 \cdot 2}{2 \cdot 2}

- \frac{1}{2}

\frac{2}{4}

\frac{1 \cdot 2}{2 \cdot 2}

\frac{1}{2}

\ln (\frac{\sqrt{| \frac{1 + 2 \cdot 2}{2} |}}{\sqrt{| \frac{2 \cdot 2 - 1}{2} |}}) - \ln (\frac{\sqrt{| - \frac{1}{2} + 2 |}}{\sqrt{| 2 + \frac{1}{2} |}})

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 2

4

■To add fractions, put them over the same denominator, then add the numerators:

- \frac{1}{2} + 2

\frac{- 1 + 2 \cdot 2}{2}

2 + \frac{1}{2}

\frac{2 \cdot 2 + 1}{2}

\ln (\frac{\sqrt{| \frac{1 + 4}{2} |}}{\sqrt{| \frac{4 - 1}{2} |}}) - \ln (\frac{\sqrt{| \frac{- 1 + 2 \cdot 2}{2} |}}{\sqrt{| \frac{2 \cdot 2 + 1}{2} |}})

ⓘ The numerator and the denominator are computed before the division:

1 + 4

5

4 - 1

3

■Multiplications are done before additions and subtractions:

2 \cdot 2

4

\ln (\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}) - \ln (\frac{\sqrt{| \frac{- 1 + 4}{2} |}}{\sqrt{| \frac{4 + 1}{2} |}})

ⓘ The numerator and the denominator are computed before the division:

- 1 + 4

3

4 + 1

5

\ln (\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}) - \ln (\frac{\sqrt{| \frac{3}{2} |}}{\sqrt{| \frac{5}{2} |}})

ⓘ For strictly positive arguments, a sum of logarithms is the logarithm of the product, a difference that of the quotient, ln(a) + ln(b) = ln(a*b):

\ln (\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}) - \ln (\frac{\sqrt{| \frac{3}{2} |}}{\sqrt{| \frac{5}{2} |}})

\ln (\frac{\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}}{\frac{\sqrt{| \frac{3}{2} |}}{\sqrt{| \frac{5}{2} |}}})

\ln (\frac{\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}}{\frac{\sqrt{| \frac{3}{2} |}}{\sqrt{| \frac{5}{2} |}}})

ⓘ Dividing is multiplying by the reciprocal:

\frac{\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}}{\frac{\sqrt{| \frac{3}{2} |}}{\sqrt{| \frac{5}{2} |}}}

\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}} \cdot \frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}

■Multiplying powers with the same base adds the exponents:

\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}} \cdot \frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}}

{(\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}})}^{2}

\ln ({(\frac{\sqrt{| \frac{5}{2} |}}{\sqrt{| \frac{3}{2} |}})}^{2})

ⓘ The expression inside the absolute value is nonnegative, so its absolute value equals that expression:

| \frac{5}{2} |

\frac{5}{2}

| \frac{3}{2} |

\frac{3}{2}

\ln ({(\frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{3}{2}}})}^{2})

ⓘ The root of a quotient is the quotient of the roots, sqrt(a/b) = sqrt(a)/sqrt(b):

\sqrt{\frac{5}{2}}

\frac{\sqrt{5}}{\sqrt{2}}

\sqrt{\frac{3}{2}}

\frac{\sqrt{3}}{\sqrt{2}}

■The power of a quotient is the quotient of the powers, (a/b)^n = a^n/b^n:

{(\frac{\frac{\sqrt{5}}{\sqrt{2}}}{\frac{\sqrt{3}}{\sqrt{2}}})}^{2}

\frac{{(\frac{\sqrt{5}}{\sqrt{2}})}^{2}}{{(\frac{\sqrt{3}}{\sqrt{2}})}^{2}}

\ln (\frac{{(\frac{\sqrt{5}}{\sqrt{2}})}^{2}}{{(\frac{\sqrt{3}}{\sqrt{2}})}^{2}})

ⓘ The power of a quotient is the quotient of the powers, (a/b)^n = a^n/b^n:

{(\frac{\sqrt{5}}{\sqrt{2}})}^{2}

\frac{{(\sqrt{5})}^{2}}{{(\sqrt{2})}^{2}}

{(\frac{\sqrt{3}}{\sqrt{2}})}^{2}

\frac{{(\sqrt{3})}^{2}}{{(\sqrt{2})}^{2}}

\ln (\frac{\frac{{(\sqrt{5})}^{2}}{{(\sqrt{2})}^{2}}}{\frac{{(\sqrt{3})}^{2}}{{(\sqrt{2})}^{2}}})

ⓘ Dividing is multiplying by the reciprocal:

\frac{\frac{{(\sqrt{5})}^{2}}{{(\sqrt{2})}^{2}}}{\frac{{(\sqrt{3})}^{2}}{{(\sqrt{2})}^{2}}}

\frac{{(\sqrt{5})}^{2}}{{(\sqrt{2})}^{2}} \cdot \frac{{(\sqrt{2})}^{2}}{{(\sqrt{3})}^{2}}

■A power of a power multiplies the exponents, (a^n)^m = a^(n*m):

{(\sqrt{5})}^{2}

5^{\frac{1}{2} \cdot 2}

{(\sqrt{2})}^{2}

2^{\frac{1}{2} \cdot 2}

{(\sqrt{3})}^{2}

3^{\frac{1}{2} \cdot 2}

\ln (\frac{5^{\frac{1}{2} \cdot 2}}{2^{\frac{1}{2} \cdot 2}} \cdot \frac{2^{\frac{1}{2} \cdot 2}}{3^{\frac{1}{2} \cdot 2}})

ⓘ Multiplying a fraction by a number multiplies its numerator:

\frac{1}{2} \cdot 2

\frac{1 \cdot 2}{2}

\frac{2}{2}

■To multiply fractions, multiply the numerators together and the denominators together:

\frac{5^{\frac{2}{2}}}{2^{\frac{2}{2}}} \cdot \frac{2^{\frac{2}{2}}}{3^{\frac{2}{2}}}

\frac{5^{\frac{2}{2}} \cdot 2^{\frac{2}{2}}}{2^{\frac{2}{2}} \cdot 3^{\frac{2}{2}}}

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor

2^{\frac{2}{2}}

\frac{5^{\frac{2}{2}} \cdot 2^{\frac{2}{2}}}{2^{\frac{2}{2}} \cdot 3^{\frac{2}{2}}}

\frac{5^{\frac{2}{2}}}{3^{\frac{2}{2}}}

\ln (\frac{5^{\frac{2}{2}}}{3^{\frac{2}{2}}})

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

\ln (\frac{5^{1}}{3^{1}})

▶Integral:

\ln (\frac{5}{3})

9) The numeric value of:

\ln (\frac{5}{3})

equals:

\ln (\approx 1.6667)

▷

\approx 0.5108