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

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

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integral from 2 to 3 of (x^2)/(x^2-1)

\int_{2}^{3} \frac{x ^{2}}{x ^{2} - 1} d x

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

Decimal

1.20273 \dots

Solution steps

\int_{2}^{3} \frac{x ^{2}}{x ^{2} - 1} d x

Factor

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

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

Take the constant out:

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

= - \int_{2}^{3} \frac{x ^{2}}{- x ^{2} + 1} d x

\frac{x ^{2}}{- x ^{2} + 1} = \frac{1}{- x ^{2} + 1} - 1

= - \int_{2}^{3} \frac{1}{- x ^{2} + 1} - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{2}^{3} \frac{1}{- x ^{2} + 1} d x = \frac{ln ( 2 ) - ln ( 3 )}{2}

\int_{2}^{3} 1 d x = 1

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

Simplify

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

\int_{0}^{4} x^{2} - 9 d x

\int_{0}^{1} 4 x^{2} + 1 x d x

\int_{4}^{5} 16 - x^{2} d x

\int_{0}^{10} \frac{1}{1 + x ^{2}} d x

\int_{0}^{1} x^{2} + x^{4} d x

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