What is the integral of 3/(2t-5) ?

The integral of 3/(2t-5) is 3/2 ln|2t-5|+C

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integral of 3/(2t-5)

\int \frac{3}{2 t - 5} d t

\frac{3}{2} ln ∣ 2 t - 5 ∣ + C

Solution steps

\int \frac{3}{2 t - 5} d t

Take the constant out:

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

= 3 \cdot \int \frac{1}{2 t - 5} d t

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 t - 5

= 3 \cdot \frac{1}{2} ln ∣ 2 t - 5 ∣

Simplify

3 \cdot \frac{1}{2} ln ∣ 2 t - 5 ∣ : \frac{3}{2} ln ∣ 2 t - 5 ∣

= \frac{3}{2} ln ∣ 2 t - 5 ∣

Add a constant to the solution

= \frac{3}{2} ln ∣ 2 t - 5 ∣ + C

\int \frac{x}{16 - x ^{4}} d x

\frac{\partial}{\partial x} (x - x y + 3)

\int_{4}^{9} \frac{t - 3}{t} d t

x \to 0 lim (cos^{2} (5 x))

x \to - \infty lim (6)