What is the integral of (3x-7)^3 ?

The integral of (3x-7)^3 is 1/12 (3x-7)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x-7)^3

\int (3 x - 7)^{3} d x

\frac{1}{12} (3 x - 7)^{4} + C

Solution steps

\int (3 x - 7)^{3} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x - 7

= \frac{1}{3} \cdot \frac{( 3 x - 7 ) ^{4}}{4}

Simplify

\frac{1}{3} \cdot \frac{( 3 x - 7 ) ^{4}}{4} : \frac{1}{12} (3 x - 7)^{4}

= \frac{1}{12} (3 x - 7)^{4}

Add a constant to the solution

= \frac{1}{12} (3 x - 7)^{4} + C

\int tan^{6} (5 x) d x

\int y^{4} e^{- y} d y

\int \frac{5}{3 a d} d y

\int \frac{1}{x ^{2} - 8} d x

\int 24 t - t^{2} + 5 t^{\frac{1}{2}} d t