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

The integral of 2x(x^2+1)^2 is (x^6)/3+x^4+x^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{x ^{6}}{3} + x^{4} + x^{2} + C

Solution steps

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

Take the constant out:

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

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

Expand

x (x^{2} + 1)^{2} : x^{5} + 2 x^{3} + x

= 2 \cdot \int x^{5} + 2 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

= 2 (\int x^{5} d x + \int 2 x^{3} d x + \int x d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

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

Simplify

2 (\frac{x ^{6}}{6} + \frac{x ^{4}}{2} + \frac{x ^{2}}{2}) : \frac{x ^{6}}{3} + x^{4} + x^{2}

= \frac{x ^{6}}{3} + x^{4} + x^{2}

Add a constant to the solution

= \frac{x ^{6}}{3} + x^{4} + x^{2} + C

What is the integral of 2x(x^2+1)^2 ?
The integral of 2x(x^2+1)^2 is (x^6)/3+x^4+x^2+C