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

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

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

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

- \frac{1}{2} ln ∣ 2 t + 1 ∣ + 2 ln ∣ t + 2 ∣ + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{t}{2 t ^{2} + 5 t + 2} d t

Take the partial fraction of

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

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

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{1}{3 ( 2 t + 1 )} d t + \int \frac{2}{3 ( t + 2 )} d t)

\int \frac{1}{3 ( 2 t + 1 )} d t = \frac{1}{6} ln ∣ 2 t + 1 ∣

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

= 3 (- \frac{1}{6} ln ∣ 2 t + 1 ∣ + \frac{2}{3} ln ∣ t + 2 ∣)

Simplify

3 (- \frac{1}{6} ln ∣ 2 t + 1 ∣ + \frac{2}{3} ln ∣ t + 2 ∣) : - \frac{1}{2} ln ∣ 2 t + 1 ∣ + 2 ln ∣ t + 2 ∣

= - \frac{1}{2} ln ∣ 2 t + 1 ∣ + 2 ln ∣ t + 2 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 2 t + 1 ∣ + 2 ln ∣ t + 2 ∣ + C

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