What is the integral from 4 to 7 of 1/(x^2-1) ?

The integral from 4 to 7 of 1/(x^2-1) is -(ln(4/3)-ln(5)+ln(3))/2

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integral from 4 to 7 of 1/(x^2-1)

\int_{4}^{7} \frac{1}{x ^{2} - 1} d x

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

Decimal Notation

0.11157 \dots

Solution steps

\int_{4}^{7} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{4}^{7} \frac{1}{- ( - x ^{2} + 1 )} d x

Take the constant out:

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

= - \int_{4}^{7} \frac{1}{- x ^{2} + 1} d x

Use the common integral:

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

= - [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{7}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{7} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{7}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{7}

Compute the boundaries:

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

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

Simplify

= - \frac{ln ( \frac{4}{3} ) - ln ( 5 ) + ln ( 3 )}{2}

\frac{d y}{d x} = 6 x + 2

\frac{\partial}{\partial y} (x^{5} + x y^{8} + 6)

y^{^{'}} = x (y + 1)

\int \frac{1}{25 + 3 x ^{2}} d x

\int \frac{1}{x ^{2} 16 x ^{2} - 9} d x

What is the integral from 4 to 7 of 1/(x^2-1) ?
The integral from 4 to 7 of 1/(x^2-1) is -(ln(4/3)-ln(5)+ln(3))/2