What is the integral of (9t^2-2)/(3t+1) ?

The integral of (9t^2-2)/(3t+1) is 3/2 t^2-t-1/3 ln|3t+1|-1/2+C

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integral of (9t^2-2)/(3t+1)

\int \frac{9 t ^{2} - 2}{3 t + 1} d t

\frac{3}{2} t^{2} - t - \frac{1}{3} ln ∣ 3 t + 1 ∣ - \frac{1}{2} + C

Solution steps

\int \frac{9 t ^{2} - 2}{3 t + 1} d t

Expand

\frac{9 t ^{2} - 2}{3 t + 1} : \frac{9 t ^{2}}{3 t + 1} - \frac{2}{3 t + 1}

Apply the Sum Rule:

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

= \int \frac{9 t ^{2}}{3 t + 1} d t - \int \frac{2}{3 t + 1} d t

\int \frac{9 t ^{2}}{3 t + 1} d t = \frac{1}{6} (9 t^{2} - 6 t + 2 ln ∣ 3 t + 1 ∣ - 3)

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

= \frac{1}{6} (9 t^{2} - 6 t + 2 ln ∣ 3 t + 1 ∣ - 3) - \frac{2}{3} ln ∣ 3 t + 1 ∣

Simplify

\frac{1}{6} (9 t^{2} - 6 t + 2 ln ∣ 3 t + 1 ∣ - 3) - \frac{2}{3} ln ∣ 3 t + 1 ∣ : \frac{3}{2} t^{2} - t - \frac{1}{3} ln ∣ 3 t + 1 ∣ - \frac{1}{2}

= \frac{3}{2} t^{2} - t - \frac{1}{3} ln ∣ 3 t + 1 ∣ - \frac{1}{2}

Add a constant to the solution

= \frac{3}{2} t^{2} - t - \frac{1}{3} ln ∣ 3 t + 1 ∣ - \frac{1}{2} + C

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What is the integral of (9t^2-2)/(3t+1) ?
The integral of (9t^2-2)/(3t+1) is 3/2 t^2-t-1/3 ln|3t+1|-1/2+C