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

\int 5 e^{- 5 x} d x

\frac{d}{d x} (arctan (\frac{1}{1 + x ^{2}}))

\int_{- 5}^{5} 4 (25 - x^{2}) d x

8 x - 5

\frac{d ^{4}}{d x ^{4}} (ln (1 + x))

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