What is the integral from 0 to 8 of 9/(10x^2+11x+1) ?

The integral from 0 to 8 of 9/(10x^2+11x+1) is 2ln(3)

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integral from 0 to 8 of 9/(10x^2+11x+1)

\int_{0}^{8} \frac{9}{10 x ^{2} + 11 x + 1} d x

2 ln (3)

Decimal Notation

2.19722 \dots

Solution steps

\int_{0}^{8} \frac{9}{10 x ^{2} + 11 x + 1} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{8} \frac{1}{10 x ^{2} + 11 x + 1} d x

Complete the square

10 x^{2} + 11 x + 1 : 10 (x + \frac{11}{20})^{2} - \frac{81}{40}

= 9 \cdot \int_{0}^{8} \frac{1}{10 ( x + \frac{11}{20} ) ^{2} - \frac{81}{40}} d x

Apply u-substitution

: \int_{\frac{11}{20}}^{\frac{171}{20}} \frac{40}{400 u ^{2} - 81} d u

= 9 \cdot \int_{\frac{11}{20}}^{\frac{171}{20}} \frac{40}{400 u ^{2} - 81} d u

Take the constant out:

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

= 9 \cdot 40 \cdot \int_{\frac{11}{20}}^{\frac{171}{20}} \frac{1}{400 u ^{2} - 81} d u

Apply Integral Substitution

= 9 \cdot 40 \cdot \int_{\frac{11}{9}}^{19} \frac{1}{180 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 9 \cdot 40 \cdot \frac{1}{180} \cdot \int_{\frac{11}{9}}^{19} \frac{1}{v ^{2} - 1} d v

Factor

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

= 9 \cdot 40 \cdot \frac{1}{180} \cdot \int_{\frac{11}{9}}^{19} \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 9 \cdot 40 \cdot \frac{1}{180} (- \int_{\frac{11}{9}}^{19} \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}

= 9 \cdot 40 \cdot \frac{1}{180} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{11}{9}}^{19})

Simplify

9 \cdot 40 \cdot \frac{1}{180} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{11}{9}}^{19}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{11}{9}}^{19}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{11}{9}}^{19}

Compute the boundaries:

\frac{ln ( 20 ) - ln ( 18 ) - ln ( \frac{20}{9} ) + ln ( \frac{2}{9} )}{2}

= - 2 \cdot \frac{ln ( 20 ) - ln ( 18 ) - ln ( \frac{20}{9} ) + ln ( \frac{2}{9} )}{2}

Simplify

= 2 ln (3)

\frac{d y}{d x} = \frac{x ^{2}}{6 e ^{y}}

f (x) = 4 x^{2} - 7 x

3 t sin (t)

\frac{\partial}{\partial x} (\frac{5 y}{x ^{5}})

x y^{^{'}} - 3 y^{2} = 0

What is the integral from 0 to 8 of 9/(10x^2+11x+1) ?
The integral from 0 to 8 of 9/(10x^2+11x+1) is 2ln(3)