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

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

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

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

- ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣ + C

Solution steps

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

Complete the square

2 x^{2} + 3 x + 1 : 2 (x + \frac{3}{4})^{2} - \frac{1}{8}

= \int \frac{1}{2 ( x + \frac{3}{4} ) ^{2} - \frac{1}{8}} d x

Apply u-substitution

: \int \frac{8}{16 u ^{2} - 1} d u

= \int \frac{8}{16 u ^{2} - 1} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{16 u ^{2} - 1} d u

Factor

16 u^{2} - 1 : - (- 16 u^{2} + 1)

= 8 \cdot \int \frac{1}{- ( - 16 u ^{2} + 1 )} d u

Take the constant out:

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

= 8 (- \int \frac{1}{- 16 u ^{2} + 1} d u)

Apply Integral Substitution

= 8 (- \int \frac{1}{4 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 8 (- \frac{1}{4} \cdot \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}

= 8 (- \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 8 (- \frac{1}{4} (\frac{ln 4 ( x + \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x + \frac{3}{4} ) - 1}{2}))

Simplify

8 (- \frac{1}{4} (\frac{ln 4 ( x + \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x + \frac{3}{4} ) - 1}{2})) : - ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣

= - ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣

Add a constant to the solution

= - ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣ + C

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