What is the integral of 7x(3x^2-2)^6 ?

The integral of 7x(3x^2-2)^6 is 7((729x^{14})/(14)-243x^{12}+486x^{10}-540x^8+360x^6-144x^4+32x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 7x(3x^2-2)^6

\int 7 x (3 x^{2} - 2)^{6} d x

7 (\frac{729 x ^{14}}{14} - 243 x^{12} + 486 x^{10} - 540 x^{8} + 360 x^{6} - 144 x^{4} + 32 x^{2}) + C

Solution steps

\int 7 x (3 x^{2} - 2)^{6} d x

Take the constant out:

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

= 7 \cdot \int x (3 x^{2} - 2)^{6} d x

Expand

x (3 x^{2} - 2)^{6} : 729 x^{13} - 2916 x^{11} + 4860 x^{9} - 4320 x^{7} + 2160 x^{5} - 576 x^{3} + 64 x

= 7 \cdot \int 729 x^{13} - 2916 x^{11} + 4860 x^{9} - 4320 x^{7} + 2160 x^{5} - 576 x^{3} + 64 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 (\int 729 x^{13} d x - \int 2916 x^{11} d x + \int 4860 x^{9} d x - \int 4320 x^{7} d x + \int 2160 x^{5} d x - \int 576 x^{3} d x + \int 64 x d x)

\int 729 x^{13} d x = \frac{729 x ^{14}}{14}

\int 2916 x^{11} d x = 243 x^{12}

\int 4860 x^{9} d x = 486 x^{10}

\int 4320 x^{7} d x = 540 x^{8}

\int 2160 x^{5} d x = 360 x^{6}

\int 576 x^{3} d x = 144 x^{4}

\int 64 x d x = 32 x^{2}

= 7 (\frac{729 x ^{14}}{14} - 243 x^{12} + 486 x^{10} - 540 x^{8} + 360 x^{6} - 144 x^{4} + 32 x^{2})

Add a constant to the solution

= 7 (\frac{729 x ^{14}}{14} - 243 x^{12} + 486 x^{10} - 540 x^{8} + 360 x^{6} - 144 x^{4} + 32 x^{2}) + C

What is the integral of 7x(3x^2-2)^6 ?
The integral of 7x(3x^2-2)^6 is 7((729x^{14})/(14)-243x^{12}+486x^{10}-540x^8+360x^6-144x^4+32x^2)+C