What is the integral from 2 to 3 of 1/(-7+16x-4x^2) ?

The integral from 2 to 3 of 1/(-7+16x-4x^2) is (ln(5))/(12)

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integral from 2 to 3 of 1/(-7+16x-4x^2)

\int_{2}^{3} \frac{1}{- 7 + 16 x - 4 x ^{2}} d x

\frac{ln ( 5 )}{12}

Decimal Notation

0.13411 \dots

Solution steps

\int_{2}^{3} \frac{1}{- 7 + 16 x - 4 x ^{2}} d x

Complete the square

- 7 + 16 x - 4 x^{2} : - 4 (x - 2)^{2} + 9

= \int_{2}^{3} \frac{1}{- 4 ( x - 2 ) ^{2} + 9} d x

Apply u-substitution

: \int_{0}^{1} \frac{1}{- 4 u ^{2} + 9} d u

= \int_{0}^{1} \frac{1}{- 4 u ^{2} + 9} d u

Apply Integral Substitution

= \int_{0}^{\frac{2}{3}} \frac{1}{6 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{\frac{2}{3}} \frac{1}{- v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{6} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{2}{3}}

Simplify

\frac{1}{6} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{2}{3}} : \frac{1}{6} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{2}{3}}

= \frac{1}{6} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{2}{3}}

Compute the boundaries:

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

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

Simplify

= \frac{ln ( 5 )}{12}

\int 1 0^{x^{2}} x d x

x \to - \infty lim (x^{3} e^{x})

x \to - 2 - lim (f (x) x)

x \to 0 lim (\frac{2}{x ^{3} + x})

y = π

What is the integral from 2 to 3 of 1/(-7+16x-4x^2) ?
The integral from 2 to 3 of 1/(-7+16x-4x^2) is (ln(5))/(12)