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

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

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

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

- 4 ln (\frac{3}{4})

Decimal

1.15072 \dots

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

: \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{4}{4 u ^{2} - 1} d u

= 4 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{4}{4 u ^{2} - 1} d u

Take the constant out:

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

= 4 \cdot 4 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{4 u ^{2} - 1} d u

Factor

4 u^{2} - 1 : - (- 4 u^{2} + 1)

= 4 \cdot 4 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{- ( - 4 u ^{2} + 1 )} d u

Take the constant out:

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

= 4 \cdot 4 (- \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{- 4 u ^{2} + 1} d u)

Apply Integral Substitution

= 4 \cdot 4 (- \int_{3}^{5} \frac{1}{2 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

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

= 4 \cdot 4 (- \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{3}^{5})

Simplify

4 \cdot 4 (- \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{3}^{5}) : - 8 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{3}^{5}

= - 8 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{3}^{5}

Compute the boundaries:

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

= - 8 \cdot \frac{ln ( 6 ) - 3 ln ( 2 )}{2}

Simplify

= - 4 ln (\frac{3}{4})

\frac{d}{d x} (x^{\frac{2}{5}} (x - 5))

y^{^{''}} - 6 y^{^{'}} = 36 t, y (0) = 6, y^{^{'}} (0) = 0

\int_{0}^{8} x^{3} e^{x} d x

f (x) = 3 x^{\frac{1}{2}} + x^{- \frac{3}{2}}

\frac{x - 4}{x + 4}

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