What is the integral of (2u-1)/(4u^2-1) ?

The integral of (2u-1)/(4u^2-1) is 1/4 ln|4u^2-1|+1/4 (ln|2u+1|-ln|2u-1|)+C

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integral of (2u-1)/(4u^2-1)

\int \frac{2 u - 1}{4 u ^{2} - 1} d u

\frac{1}{4} ln 4 u^{2} - 1 + \frac{1}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣) + C

Solution steps

\int \frac{2 u - 1}{4 u ^{2} - 1} d u

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{4 y}{1} d y

y = 2 e^{x} cos (x), (0, 2)

\int x^{π} d x

\int 2 x + 3 d x

x \to 0 lim (x^{x^{x}} - 1)

What is the integral of (2u-1)/(4u^2-1) ?
The integral of (2u-1)/(4u^2-1) is 1/4 ln|4u^2-1|+1/4 (ln|2u+1|-ln|2u-1|)+C