What is the integral of (2(2t-1))/(t(3t^2+4t-1)) ?

The integral of (2(2t-1))/(t(3t^2+4t-1)) is 2ln|t|-ln|9t^2+12t-3|+C

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integral of (2(2t-1))/(t(3t^2+4t-1))

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

2 ln ∣ t ∣ - ln 9 t^{2} + 12 t - 3 + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 2 \cdot \int \frac{1}{t} + \frac{- 3 t - 2}{3 t ^{2} + 4 t - 1} 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

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

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{- 3 t - 2}{3 t ^{2} + 4 t - 1} d t = - \frac{1}{2} ln 9 t^{2} + 12 t - 3

= 2 (ln ∣ t ∣ - \frac{1}{2} ln 9 t^{2} + 12 t - 3)

Simplify

2 (ln ∣ t ∣ - \frac{1}{2} ln 9 t^{2} + 12 t - 3) : 2 ln ∣ t ∣ - ln 9 t^{2} + 12 t - 3

= 2 ln ∣ t ∣ - ln 9 t^{2} + 12 t - 3

Add a constant to the solution

= 2 ln ∣ t ∣ - ln 9 t^{2} + 12 t - 3 + C

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

x e^{(x + 2)}

2 x^{2} ln (x), 8 ln (x)

\int \frac{( x ^{3} )}{3} d x

y = 3 x^{2} + 17 x + 10

What is the integral of (2(2t-1))/(t(3t^2+4t-1)) ?
The integral of (2(2t-1))/(t(3t^2+4t-1)) is 2ln|t|-ln|9t^2+12t-3|+C