What is the integral from 0 to 2 of 3/(4x^2+5x+1) ?

The integral from 0 to 2 of 3/(4x^2+5x+1) is ln(3)

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integral from 0 to 2 of 3/(4x^2+5x+1)

\int_{0}^{2} \frac{3}{4 x ^{2} + 5 x + 1} d x

ln (3)

Decimal

1.09861 \dots

Solution steps

\int_{0}^{2} \frac{3}{4 x ^{2} + 5 x + 1} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{2} \frac{1}{4 x ^{2} + 5 x + 1} d x

Complete the square

4 x^{2} + 5 x + 1 : 4 (x + \frac{5}{8})^{2} - \frac{9}{16}

= 3 \cdot \int_{0}^{2} \frac{1}{4 ( x + \frac{5}{8} ) ^{2} - \frac{9}{16}} d x

Apply u-substitution

: \int_{\frac{5}{8}}^{\frac{21}{8}} \frac{16}{64 u ^{2} - 9} d u

= 3 \cdot \int_{\frac{5}{8}}^{\frac{21}{8}} \frac{16}{64 u ^{2} - 9} d u

Take the constant out:

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

= 3 \cdot 16 \cdot \int_{\frac{5}{8}}^{\frac{21}{8}} \frac{1}{64 u ^{2} - 9} d u

Apply Integral Substitution

= 3 \cdot 16 \cdot \int_{\frac{5}{3}}^{7} \frac{1}{24 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 3 \cdot 16 \cdot \frac{1}{24} \cdot \int_{\frac{5}{3}}^{7} \frac{1}{v ^{2} - 1} d v

Factor

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

= 3 \cdot 16 \cdot \frac{1}{24} \cdot \int_{\frac{5}{3}}^{7} \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

= 3 \cdot 16 \cdot \frac{1}{24} (- \int_{\frac{5}{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}

= 3 \cdot 16 \cdot \frac{1}{24} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{7})

Simplify

3 \cdot 16 \cdot \frac{1}{24} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{7}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{5}{3}}^{7}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{5}{3}}^{7}

Compute the boundaries:

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

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

Simplify

= ln (3)

\frac{\partial}{\partial x} (f (x))

\int \frac{1}{( x - 1 ) ^{3}} d x

\frac{d}{d x} (\frac{x ^{3}}{11})

\int sin^{5} (x) \cdot cos (x) d x

\frac{d}{d x} (\frac{sin ( π ) x}{ln ( x )})

What is the integral from 0 to 2 of 3/(4x^2+5x+1) ?
The integral from 0 to 2 of 3/(4x^2+5x+1) is ln(3)