What is the integral of (x+3)(x^2+1) ?

The integral of (x+3)(x^2+1) is (x^4)/4+(x^2)/2+x^3+3x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x+3)(x^2+1)

\int (x + 3) (x^{2} + 1) d x

\frac{x ^{4}}{4} + \frac{x ^{2}}{2} + x^{3} + 3 x + C

Solution steps

\int (x + 3) (x^{2} + 1) d x

Expand

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

= \int x^{3} + x + 3 x^{2} + 3 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

= \int x^{3} d x + \int x d x + \int 3 x^{2} d x + \int 3 d x

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

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

\int 3 x^{2} d x = x^{3}

\int 3 d x = 3 x

= \frac{x ^{4}}{4} + \frac{x ^{2}}{2} + x^{3} + 3 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{x ^{2}}{2} + x^{3} + 3 x + C

What is the integral of (x+3)(x^2+1) ?
The integral of (x+3)(x^2+1) is (x^4)/4+(x^2)/2+x^3+3x+C