What is the integral of ((4x-1))/((2x+1)) ?

The integral of ((4x-1))/((2x+1)) is 2x+1-3/2 ln|2x+1|+C

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integral of ((4x-1))/((2x+1))

\int \frac{( 4 x - 1 )}{( 2 x + 1 )} d x

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

Solution steps

\int \frac{4 x - 1}{2 x + 1} d x

Apply u-substitution

: \int \frac{2 u - 3}{2 u} d u

= \int \frac{2 u - 3}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{2 u - 3}{u} d u

Expand

\frac{2 u - 3}{u} : 2 - \frac{3}{u}

= \frac{1}{2} \cdot \int 2 - \frac{3}{u} d u

Apply the Sum Rule:

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

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

\int 2 d u = 2 u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= \frac{1}{2} (2 u - 3 ln ∣ u ∣)

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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

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What is the integral of ((4x-1))/((2x+1)) ?
The integral of ((4x-1))/((2x+1)) is 2x+1-3/2 ln|2x+1|+C