What is the integral of (3/(-2t+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3/(-2t+1))

\int (\frac{3}{- 2 t + 1}) d t

- \frac{3}{2} ln ∣ - 2 t + 1 ∣ + C

Solution steps

\int \frac{3}{- 2 t + 1} 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 + 1} 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 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = - 2 t + 1

= 3 (- \frac{1}{2} ln ∣ - 2 t + 1 ∣)

Simplify

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

= - \frac{3}{2} ln ∣ - 2 t + 1 ∣

Add a constant to the solution

= - \frac{3}{2} ln ∣ - 2 t + 1 ∣ + C

\int \frac{2 sec ^{2} ( x )}{( 1 + tan ( x ) ) ^{2}} d x

\int (- 9 cos (t)) d t

\int y^{2} + 4 y d y

\int sin^{3} (r) cos^{6} (r) d r

\int e^{- 2 x + 18} d x

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