What is the integral from 0 to 3 of 4/(5x^2+6x+1) ?

The integral from 0 to 3 of 4/(5x^2+6x+1) is 2ln(2)

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integral from 0 to 3 of 4/(5x^2+6x+1)

\int_{0}^{3} \frac{4}{5 x ^{2} + 6 x + 1} d x

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

\int_{0}^{3} \frac{4}{5 x ^{2} + 6 x + 1} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{3} \frac{1}{5 x ^{2} + 6 x + 1} d x

Complete the square

5 x^{2} + 6 x + 1 : 5 (x + \frac{3}{5})^{2} - \frac{4}{5}

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

Apply u-substitution

: \int_{\frac{3}{5}}^{\frac{18}{5}} \frac{5}{25 u ^{2} - 4} d u

= 4 \cdot \int_{\frac{3}{5}}^{\frac{18}{5}} \frac{5}{25 u ^{2} - 4} d u

Take the constant out:

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

= 4 \cdot 5 \cdot \int_{\frac{3}{5}}^{\frac{18}{5}} \frac{1}{25 u ^{2} - 4} d u

Apply Integral Substitution

= 4 \cdot 5 \cdot \int_{\frac{3}{2}}^{9} \frac{1}{10 ( 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 5 \cdot \frac{1}{10} \cdot \int_{\frac{3}{2}}^{9} \frac{1}{v ^{2} - 1} d v

Factor

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

= 4 \cdot 5 \cdot \frac{1}{10} \cdot \int_{\frac{3}{2}}^{9} \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

= 4 \cdot 5 \cdot \frac{1}{10} (- \int_{\frac{3}{2}}^{9} \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 5 \cdot \frac{1}{10} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{3}{2}}^{9})

Simplify

4 \cdot 5 \cdot \frac{1}{10} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{3}{2}}^{9}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{3}{2}}^{9}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{3}{2}}^{9}

Compute the boundaries:

\frac{- 4 ln ( 2 ) + ln ( 10 ) - ln ( \frac{5}{2} )}{2}

= - 2 \cdot \frac{- 4 ln ( 2 ) + ln ( 10 ) - ln ( \frac{5}{2} )}{2}

Simplify

= 2 ln (2)

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

\frac{d}{d x} (9 x + 2)

y (1 - t) d t + t^{2} (1 - t) d y = 0

\frac{d}{d x} (- \frac{2 x}{( x ^{2} + 1 ) ^{2}})

f (x) = 4 x^{2} - 3 x, at x = - 1

What is the integral from 0 to 3 of 4/(5x^2+6x+1) ?
The integral from 0 to 3 of 4/(5x^2+6x+1) is 2ln(2)