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

The integral from-2 to 2 of 1/(x^2-9) is -1/3 ln(5)

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

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

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

Decimal Notation

- 0.53647 \dots

Solution steps

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

Apply Integral Substitution

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

Factor

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

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

Simplify

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

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

Simplify

= - \frac{1}{3} ln (5)

\int tan^{4} (x se) c^{2} x d x

\int \frac{( 3 + ln ( 2 x ) ) ^{3}}{x} d x

\frac{d v}{d t} = 9.8 - v

f (x) = x e^{2 x^{2} - 4 x}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 15 x + 5 e^{- x}

What is the integral from-2 to 2 of 1/(x^2-9) ?
The integral from-2 to 2 of 1/(x^2-9) is -1/3 ln(5)