What is the integral from-1 to 1 of 1/(4x^2-9) ?

The integral from-1 to 1 of 1/(4x^2-9) is -1/6 ln(5)

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integral from-1 to 1 of 1/(4x^2-9)

\int_{- 1}^{1} \frac{1}{4 x ^{2} - 9} d x

- \frac{1}{6} ln (5)

Decimal

- 0.26823 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{4 x ^{2} - 9} d x

Apply Integral Substitution

= \int_{- \frac{2}{3}}^{\frac{2}{3}} \frac{1}{6 ( u ^{2} - 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{- \frac{2}{3}}^{\frac{2}{3}} \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{6} \cdot \int_{- \frac{2}{3}}^{\frac{2}{3}} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} (- \int_{- \frac{2}{3}}^{\frac{2}{3}} \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{\frac{2}{3}})

Simplify

\frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{\frac{2}{3}}) : - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{\frac{2}{3}}

= - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{\frac{2}{3}}

Compute the boundaries:

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

= - \frac{1}{6} \cdot \frac{2 ln ( 3 ) + 2 ln ( \frac{5}{3} )}{2}

Simplify

= - \frac{1}{6} ln (5)

\int_{3}^{8} \frac{x}{1 + x} d x

\int_{2}^{6} x x - 2 d x

\int_{0}^{2 π} (sin (x)) d x

\int_{0}^{5} x d x

\int_{0}^{4} ∣ x - 1 ∣ d x

What is the integral from-1 to 1 of 1/(4x^2-9) ?
The integral from-1 to 1 of 1/(4x^2-9) is -1/6 ln(5)