What is the integral from 0 to 1 of 2/(2x^2+3x+1) ?

The integral from 0 to 1 of 2/(2x^2+3x+1) is -2ln(2/3)

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integral from 0 to 1 of 2/(2x^2+3x+1)

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

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

Decimal Notation

0.81093 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{1} \frac{1}{2 x ^{2} + 3 x + 1} d x

Complete the square

2 x^{2} + 3 x + 1 : 2 (x + \frac{3}{4})^{2} - \frac{1}{8}

= 2 \cdot \int_{0}^{1} \frac{1}{2 ( x + \frac{3}{4} ) ^{2} - \frac{1}{8}} d x

Apply u-substitution

: \int_{\frac{3}{4}}^{\frac{7}{4}} \frac{8}{16 u ^{2} - 1} d u

= 2 \cdot \int_{\frac{3}{4}}^{\frac{7}{4}} \frac{8}{16 u ^{2} - 1} d u

Take the constant out:

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

= 2 \cdot 8 \cdot \int_{\frac{3}{4}}^{\frac{7}{4}} \frac{1}{16 u ^{2} - 1} d u

Factor

16 u^{2} - 1 : - (- 16 u^{2} + 1)

= 2 \cdot 8 \cdot \int_{\frac{3}{4}}^{\frac{7}{4}} \frac{1}{- ( - 16 u ^{2} + 1 )} d u

Take the constant out:

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

= 2 \cdot 8 (- \int_{\frac{3}{4}}^{\frac{7}{4}} \frac{1}{- 16 u ^{2} + 1} d u)

Apply Integral Substitution

= 2 \cdot 8 (- \int_{3}^{7} \frac{1}{4 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 2 \cdot 8 (- \frac{1}{4} \cdot \int_{3}^{7} \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}

= 2 \cdot 8 (- \frac{1}{4} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{3}^{7})

Simplify

2 \cdot 8 (- \frac{1}{4} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{3}^{7}) : - 4 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{3}^{7}

= - 4 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{3}^{7}

Compute the boundaries:

\frac{2 ln ( 2 ) - ln ( 6 )}{2}

= - 4 \cdot \frac{2 ln ( 2 ) - ln ( 6 )}{2}

Simplify

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

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