What is the integral of x/(2x+3) ?

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

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integral of x/(2x+3)

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

\frac{1}{4} (2 x + 3 - 3 ln ∣ 2 x + 3 ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 3}{4 u} d u

= \int \frac{u - 3}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 3}{u} d u

Expand

\frac{u - 3}{u} : 1 - \frac{3}{u}

= \frac{1}{4} \cdot \int 1 - \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}{4} (\int 1 d u - \int \frac{3}{u} d u)

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= \frac{1}{4} (u - 3 ln ∣ u ∣)

Substitute back

u = 2 x + 3

= \frac{1}{4} (2 x + 3 - 3 ln ∣ 2 x + 3 ∣)

Add a constant to the solution

= \frac{1}{4} (2 x + 3 - 3 ln ∣ 2 x + 3 ∣) + C

f (x) = \frac{x}{7 x ^{2} + 1}

\frac{x + 1}{x ^{3} + x - 2}

\frac{d y}{d t} = - \frac{1}{5} y

\int_{π}^{5 π} x^{2} sin (x) d x

8 w^{4} + 3 w^{2} + 7 w

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