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

The integral of (x^2+x+1)^2 is x^5+5/2 x^4+4x^3+7/2 x^2+2x+1/2-1/40 (2x+1)^5-(x+1/2)^3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x^{5} + \frac{5}{2} x^{4} + 4 x^{3} + \frac{7}{2} x^{2} + 2 x + \frac{1}{2} - \frac{1}{40} (2 x + 1)^{5} - (x + \frac{1}{2})^{3} + C

Solution steps

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

Complete the square

x^{2} + x + 1 : (x + \frac{1}{2})^{2} + \frac{3}{4}

= \int ((x + \frac{1}{2})^{2} + \frac{3}{4})^{2} d x

Apply u-substitution

: \int (u^{2} + \frac{3}{4})^{2} d u

= \int (u^{2} + \frac{3}{4})^{2} d u

Apply Integration By Parts

: (u^{2} + \frac{3}{4})^{2} u - \int (4 u^{3} + 3 u) u d u

= (u^{2} + \frac{3}{4})^{2} u - \int (4 u^{3} + 3 u) u d u

\int (4 u^{3} + 3 u) u d u = \frac{4 u ^{5}}{5} + u^{3}

= (u^{2} + \frac{3}{4})^{2} u - (\frac{4 u ^{5}}{5} + u^{3})

Substitute back

u = x + \frac{1}{2}

= ((x + \frac{1}{2})^{2} + \frac{3}{4})^{2} (x + \frac{1}{2}) - (\frac{4 ( x + \frac{1}{2} ) ^{5}}{5} + (x + \frac{1}{2})^{3})

Simplify

((x + \frac{1}{2})^{2} + \frac{3}{4})^{2} (x + \frac{1}{2}) - (\frac{4 ( x + \frac{1}{2} ) ^{5}}{5} + (x + \frac{1}{2})^{3}) : x^{5} + \frac{5}{2} x^{4} + 4 x^{3} + \frac{7}{2} x^{2} + 2 x + \frac{1}{2} - \frac{1}{40} (2 x + 1)^{5} - (x + \frac{1}{2})^{3}

= x^{5} + \frac{5}{2} x^{4} + 4 x^{3} + \frac{7}{2} x^{2} + 2 x + \frac{1}{2} - \frac{1}{40} (2 x + 1)^{5} - (x + \frac{1}{2})^{3}

Add a constant to the solution

= x^{5} + \frac{5}{2} x^{4} + 4 x^{3} + \frac{7}{2} x^{2} + 2 x + \frac{1}{2} - \frac{1}{40} (2 x + 1)^{5} - (x + \frac{1}{2})^{3} + C

y^{^{''}} - y^{^{'}} - 56 y = 0

\frac{3 x - 1}{e ^{x}}

\frac{\partial}{\partial x} (1 - \frac{y}{x})

x (cos (x^{2}))

\frac{( x ^{2} - 4 )}{( x ^{2} - 1 )}

What is the integral of (x^2+x+1)^2 ?
The integral of (x^2+x+1)^2 is x^5+5/2 x^4+4x^3+7/2 x^2+2x+1/2-1/40 (2x+1)^5-(x+1/2)^3+C