What is the integral of 6x(x^2+1) ?

The integral of 6x(x^2+1) is 6((x^4)/4+(x^2)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6x(x^2+1)

\int 6 x (x^{2} + 1) d x

6 (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

Solution steps

\int 6 x (x^{2} + 1) d x

Take the constant out:

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

= 6 \cdot \int x (x^{2} + 1) d x

Expand

x (x^{2} + 1) : x^{3} + x

= 6 \cdot \int x^{3} + 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

= 6 (\int x^{3} d x + \int x d x)

\int x^{3} d x = \frac{x ^{4}}{4}

\int x d x = \frac{x ^{2}}{2}

= 6 (\frac{x ^{4}}{4} + \frac{x ^{2}}{2})

Add a constant to the solution

= 6 (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

\int + (cos (a x + b)) d x

\int \frac{1}{9 + 16 r ^{2}} d r

\int \frac{72}{x ^{2} - 81} d x

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

\int (\frac{9 - x ^{2}}{x}) d x

What is the integral of 6x(x^2+1) ?
The integral of 6x(x^2+1) is 6((x^4)/4+(x^2)/2)+C