What is the integral from 0 to 1 of 2/(3x^2+4x+1) ?

The integral from 0 to 1 of 2/(3x^2+4x+1) is ln(2)

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integral from 0 to 1 of 2/(3x^2+4x+1)

\int_{0}^{1} \frac{2}{3 x ^{2} + 4 x + 1} d x

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{1} \frac{2}{3 x ^{2} + 4 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}{3 x ^{2} + 4 x + 1} d x

Complete the square

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

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

Apply u-substitution

: \int_{\frac{2}{3}}^{\frac{5}{3}} \frac{3}{9 u ^{2} - 1} d u

= 2 \cdot \int_{\frac{2}{3}}^{\frac{5}{3}} \frac{3}{9 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 3 \cdot \int_{\frac{2}{3}}^{\frac{5}{3}} \frac{1}{9 u ^{2} - 1} d u

Factor

9 u^{2} - 1 : - (- 9 u^{2} + 1)

= 2 \cdot 3 \cdot \int_{\frac{2}{3}}^{\frac{5}{3}} \frac{1}{- ( - 9 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 3 (- \int_{\frac{2}{3}}^{\frac{5}{3}} \frac{1}{- 9 u ^{2} + 1} d u)

Apply Integral Substitution

= 2 \cdot 3 (- \int_{2}^{5} \frac{1}{3 ( - 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 3 (- \frac{1}{3} \cdot \int_{2}^{5} \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 3 (- \frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{2}^{5})

Simplify

2 \cdot 3 (- \frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{2}^{5}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{2}^{5}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{2}^{5}

Compute the boundaries:

\frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 3 )}{2}

= - 2 \cdot \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 3 )}{2}

Simplify

= ln (2)

x \to 0 lim (\frac{6 ^{x} - 4 ^{x}}{x})

\int - x^{2} \frac{3}{3} d x

\frac{d}{d x} (x tan (y))

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

\frac{d}{d x} (e^{i 2 x})

What is the integral from 0 to 1 of 2/(3x^2+4x+1) ?
The integral from 0 to 1 of 2/(3x^2+4x+1) is ln(2)