What is the integral of 21(x+4)^6 ?

The integral of 21(x+4)^6 is 3(x+4)^7+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 21(x+4)^6

\int 21 (x + 4)^{6} d x

3 (x + 4)^{7} + C

Solution steps

\int 21 (x + 4)^{6} d x

Take the constant out:

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

= 21 \cdot \int (x + 4)^{6} d x

Apply u-substitution

: \int u^{6} d u

= 21 \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 21 \cdot \frac{u ^{7}}{7}

Substitute back

u = x + 4

= 21 \cdot \frac{( x + 4 ) ^{7}}{7}

Simplify

21 \cdot \frac{( x + 4 ) ^{7}}{7} : 3 (x + 4)^{7}

= 3 (x + 4)^{7}

Add a constant to the solution

= 3 (x + 4)^{7} + C

\int \frac{2 x + 3}{( x + 2 ) ( x ^{2} + 5 )} d x

\int 6 cos^{4} (4 x) d x

\int (2 + x) d x

\int \frac{cos ^{3} ( x )}{csc ( x )} d x

\int \frac{1}{7 x ^{2} + 4} d x