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

The integral of (2t-1)^3 is 1/8 (2t-1)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2t-1)^3

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

\frac{1}{8} (2 t - 1)^{4} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

= \int \frac{u ^{3}}{2} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = 2 t - 1

= \frac{1}{2} \cdot \frac{( 2 t - 1 ) ^{4}}{4}

Simplify

\frac{1}{2} \cdot \frac{( 2 t - 1 ) ^{4}}{4} : \frac{1}{8} (2 t - 1)^{4}

= \frac{1}{8} (2 t - 1)^{4}

Add a constant to the solution

= \frac{1}{8} (2 t - 1)^{4} + C

\int \frac{3 x}{3 x - 1} d x

\int - 74 d x

\int - 2 e^{- 4 x} d x

\int (y^{2} - 2 x) d x