What is the integral from 4 to 5 of x/(4-x^2) ?

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

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integral from 4 to 5 of x/(4-x^2)

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

\frac{1}{2} ln (\frac{4}{7})

Decimal Notation

- 0.27980 \dots

Solution steps

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

Apply u-substitution

: \int_{- 12}^{- 21} - \frac{1}{2 u} d u

= \int_{- 12}^{- 21} - \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 21}^{- 12} - \frac{1}{2 u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{- 21}^{- 12} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{2} [ln ∣ u ∣]_{- 21}^{- 12})

Simplify

= \frac{1}{2} [ln ∣ u ∣]_{- 21}^{- 12}

Compute the boundaries:

ln (12) - ln (21)

= \frac{1}{2} (ln (12) - ln (21))

Simplify

= \frac{1}{2} ln (\frac{4}{7})

\int_{0}^{1} \frac{9 x}{( 2 x + 1 ) ^{3}} d x

\int_{0}^{\infty} x^{2} - 1 d x

\int_{1}^{2} \frac{sin ( ln ( x ) )}{x} d x

\int_{1}^{e} ln (x^{7}) d x

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

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