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

The integral of x/(2x^2+3x+1) is 2(1/8 ln|16x^2+24x+8|+3/8 (ln|4x+4|-ln|4x+2|))+C

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

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

2 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣)) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 ( 4 u - 3 )}{16 u ^{2} - 1} d u

= \int \frac{2 ( 4 u - 3 )}{16 u ^{2} - 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{4 u - 3}{16 u ^{2} - 1} d u

Expand

\frac{4 u - 3}{16 u ^{2} - 1} : \frac{4 u}{16 u ^{2} - 1} - \frac{3}{16 u ^{2} - 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{4 u}{16 u ^{2} - 1} d u - \int \frac{3}{16 u ^{2} - 1} d u)

\int \frac{4 u}{16 u ^{2} - 1} d u = \frac{1}{8} ln 16 u^{2} - 1

\int \frac{3}{16 u ^{2} - 1} d u = - \frac{3}{8} (ln ∣ 4 u + 1 ∣ - ln ∣ 4 u - 1 ∣)

= 2 (\frac{1}{8} ln 16 u^{2} - 1 - (- \frac{3}{8} (ln ∣ 4 u + 1 ∣ - ln ∣ 4 u - 1 ∣)))

Substitute back

u = x + \frac{3}{4}

= 2 (\frac{1}{8} ln 16 (x + \frac{3}{4})^{2} - 1 - (- \frac{3}{8} (ln 4 (x + \frac{3}{4}) + 1 - ln 4 (x + \frac{3}{4}) - 1)))

Simplify

2 (\frac{1}{8} ln 16 (x + \frac{3}{4})^{2} - 1 - (- \frac{3}{8} (ln 4 (x + \frac{3}{4}) + 1 - ln 4 (x + \frac{3}{4}) - 1))) : 2 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣))

= 2 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣))

Add a constant to the solution

= 2 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣)) + C

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What is the integral of x/(2x^2+3x+1) ?
The integral of x/(2x^2+3x+1) is 2(1/8 ln|16x^2+24x+8|+3/8 (ln|4x+4|-ln|4x+2|))+C