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

The integral of ((3t+4))/((2t+3)) is 3/2 t+9/4-1/4 ln|2t+3|+C

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

integral

\frac{( 3 t + 4 )}{( 2 t + 3 )}

\frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣ + C

Solution steps

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

Expand

\frac{3 t + 4}{2 t + 3} : \frac{3 t}{2 t + 3} + \frac{4}{2 t + 3}

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

\int \frac{3 t}{2 t + 3} d t = \frac{3}{4} (2 t + 3 - 3 ln ∣ 2 t + 3 ∣)

\int \frac{4}{2 t + 3} d t = 2 ln ∣ 2 t + 3 ∣

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

Simplify

\frac{3}{4} (2 t + 3 - 3 ln ∣ 2 t + 3 ∣) + 2 ln ∣ 2 t + 3 ∣ : \frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣

= \frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣

Add a constant to the solution

= \frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣ + C

\frac{1}{( y - 2 )}

\frac{1}{( x ^{2} + x - 2 )}

(4 x + 9)^{\frac{1}{2}}

(2 x + 8)^{3}

x^{2} arcsin (x)

What is the integral of ((3t+4))/((2t+3)) ?
The integral of ((3t+4))/((2t+3)) is 3/2 t+9/4-1/4 ln|2t+3|+C