What is the integral of (x+4)/(2x+3) ?

The integral of (x+4)/(2x+3) is 1/4 (2x+3+5ln|2x+3|)+C

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integral of (x+4)/(2x+3)

\int \frac{x + 4}{2 x + 3} d x

\frac{1}{4} (2 x + 3 + 5 ln ∣ 2 x + 3 ∣) + C

Solution steps

\int \frac{x + 4}{2 x + 3} d x

Apply u-substitution

: \int \frac{u + 5}{4 u} d u

= \int \frac{u + 5}{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 + 5}{u} d u

Expand

\frac{u + 5}{u} : 1 + \frac{5}{u}

= \frac{1}{4} \cdot \int 1 + \frac{5}{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{5}{u} d u)

\int 1 d u = u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= \frac{1}{4} (u + 5 ln ∣ u ∣)

Substitute back

u = 2 x + 3

= \frac{1}{4} (2 x + 3 + 5 ln ∣ 2 x + 3 ∣)

Add a constant to the solution

= \frac{1}{4} (2 x + 3 + 5 ln ∣ 2 x + 3 ∣) + C

y^{^{''}} + y = e^{x} sin (x)

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

\int 1 + 2 cos (x) + cos^{2} (x) d x

y = x^{3}, y = x^{2}

\frac{\partial}{\partial t} (t^{2} + 1)

What is the integral of (x+4)/(2x+3) ?
The integral of (x+4)/(2x+3) is 1/4 (2x+3+5ln|2x+3|)+C