What is the integral of 36(1+x^2)(7+18x+6x^3)^{10} ?

The integral of 36(1+x^2)(7+18x+6x^3)^{10} is 2/11 (6x^3+18x+7)^{11}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 36(1+x^2)(7+18x+6x^3)^{10}

\int 36 (1 + x^{2}) (7 + 18 x + 6 x^{3})^{10} d x

\frac{2}{11} (6 x^{3} + 18 x + 7)^{11} + C

Solution steps

\int 36 (1 + x^{2}) (7 + 18 x + 6 x^{3})^{10} d x

Take the constant out:

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

= 36 \cdot \int (1 + x^{2}) (7 + 18 x + 6 x^{3})^{10} d x

Apply u-substitution

: \int \frac{u ^{10}}{18} d u

= 36 \cdot \int \frac{u ^{10}}{18} d u

Take the constant out:

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

= 36 \cdot \frac{1}{18} \cdot \int u^{10} d u

Apply the Power Rule

: \frac{u ^{11}}{11}

= 36 \cdot \frac{1}{18} \cdot \frac{u ^{11}}{11}

Substitute back

u = 7 + 18 x + 6 x^{3}

= 36 \cdot \frac{1}{18} \cdot \frac{( 7 + 18 x + 6 x ^{3} ) ^{11}}{11}

Simplify

36 \cdot \frac{1}{18} \cdot \frac{( 7 + 18 x + 6 x ^{3} ) ^{11}}{11} : \frac{2}{11} (6 x^{3} + 18 x + 7)^{11}

= \frac{2}{11} (6 x^{3} + 18 x + 7)^{11}

Add a constant to the solution

= \frac{2}{11} (6 x^{3} + 18 x + 7)^{11} + C

\int cot (x) ln (csc (x)) d x

\int 5 x^{3} + 2 x^{2} - 6 x + 3 d x

\int (x^{2} - 4 x + 10) d x

\int \frac{5 v - 2}{v ^{2} - 4} d v

\int \frac{e ^{x}}{e ^{x} - 7} d x

What is the integral of 36(1+x^2)(7+18x+6x^3)^{10} ?
The integral of 36(1+x^2)(7+18x+6x^3)^{10} is 2/11 (6x^3+18x+7)^{11}+C