What is the integral of 21(y^6+2y^3+4)^3(2y^5+2y^2) ?

The integral of 21(y^6+2y^3+4)^3(2y^5+2y^2) is 7/4 (y^6+2y^3+4)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 21(y^6+2y^3+4)^3(2y^5+2y^2)

\int 21 (y^{6} + 2 y^{3} + 4)^{3} (2 y^{5} + 2 y^{2}) d y

\frac{7}{4} (y^{6} + 2 y^{3} + 4)^{4} + C

Solution steps

\int 21 (y^{6} + 2 y^{3} + 4)^{3} (2 y^{5} + 2 y^{2}) d y

Take the constant out:

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

= 21 \cdot \int (y^{6} + 2 y^{3} + 4)^{3} (2 y^{5} + 2 y^{2}) d y

Apply u-substitution

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

= 21 \cdot \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

= 21 \cdot \frac{1}{3} \cdot \int u^{3} d u

Apply the Power Rule

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

= 21 \cdot \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = y^{6} + 2 y^{3} + 4

= 21 \cdot \frac{1}{3} \cdot \frac{( y ^{6} + 2 y ^{3} + 4 ) ^{4}}{4}

Simplify

21 \cdot \frac{1}{3} \cdot \frac{( y ^{6} + 2 y ^{3} + 4 ) ^{4}}{4} : \frac{7}{4} (y^{6} + 2 y^{3} + 4)^{4}

= \frac{7}{4} (y^{6} + 2 y^{3} + 4)^{4}

Add a constant to the solution

= \frac{7}{4} (y^{6} + 2 y^{3} + 4)^{4} + C

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

\int 8 x^{8} d x

\int \frac{2}{5 + 9 x ^{2}} d x

\int 0.03 e^{0.08 x} d x

What is the integral of 21(y^6+2y^3+4)^3(2y^5+2y^2) ?
The integral of 21(y^6+2y^3+4)^3(2y^5+2y^2) is 7/4 (y^6+2y^3+4)^4+C