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

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

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

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

ln ∣ 2 x + 1 ∣ - ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= 3 \cdot \int \frac{2}{3 ( 2 x + 1 )} - \frac{1}{3 ( x + 2 )} d x

Apply the Sum Rule:

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

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

\int \frac{2}{3 ( 2 x + 1 )} d x = \frac{1}{3} ln ∣ 2 x + 1 ∣

\int \frac{1}{3 ( x + 2 )} d x = \frac{1}{3} ln ∣ x + 2 ∣

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

Simplify

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

= ln ∣ 2 x + 1 ∣ - ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ 2 x + 1 ∣ - ln ∣ x + 2 ∣ + C

4 t^{3}

\int (- 24 x^{5} - 10 x) d x

\frac{d}{d x} (ln (20 x))

\frac{x}{( 4 x - 3 ) ^{8}}

\int_{0}^{4} 2 x - x^{2} d x

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