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

The integral of 3/(4x^2+5x+1) is -ln| 8/3+8/3 x|+ln| 2/3+8/3 x|+C

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

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

- ln \frac{8}{3} + \frac{8}{3} x + ln \frac{2}{3} + \frac{8}{3} x + C

Solution steps

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

Apply u-substitution

: \int \frac{16}{64 u ^{2} - 9} d u

= 3 \cdot \int \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{1}{64 u ^{2} - 9} d u

Apply Integral Substitution

= 3 \cdot 16 \cdot \int \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{1}{v ^{2} - 1} d v

Factor

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

= 3 \cdot 16 \cdot \frac{1}{24} \cdot \int \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{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}))

Substitute back

= 3 \cdot 16 \cdot \frac{1}{24} (- (\frac{ln \frac{8}{3} ( x + \frac{5}{8} ) + 1}{2} - \frac{ln \frac{8}{3} ( x + \frac{5}{8} ) - 1}{2}))

Simplify

3 \cdot 16 \cdot \frac{1}{24} (- (\frac{ln \frac{8}{3} ( x + \frac{5}{8} ) + 1}{2} - \frac{ln \frac{8}{3} ( x + \frac{5}{8} ) - 1}{2})) : - ln \frac{8}{3} + \frac{8}{3} x + ln \frac{2}{3} + \frac{8}{3} x

= - ln \frac{8}{3} + \frac{8}{3} x + ln \frac{2}{3} + \frac{8}{3} x

Add a constant to the solution

= - ln \frac{8}{3} + \frac{8}{3} x + ln \frac{2}{3} + \frac{8}{3} x + C

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What is the integral of 3/(4x^2+5x+1) ?
The integral of 3/(4x^2+5x+1) is -ln| 8/3+8/3 x|+ln| 2/3+8/3 x|+C