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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int (u^{2} - 1)^{3} d u

= \int (u^{2} - 1)^{3} d u

Expand

(u^{2} - 1)^{3} : u^{6} - 3 u^{4} + 3 u^{2} - 1

= \int u^{6} - 3 u^{4} + 3 u^{2} - 1 d u

Apply the Sum Rule:

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

= \int u^{6} d u - \int 3 u^{4} d u + \int 3 u^{2} d u - \int 1 d u

\int u^{6} d u = \frac{u ^{7}}{7}

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

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

\int 1 d u = u

= \frac{u ^{7}}{7} - \frac{3 u ^{5}}{5} + u^{3} - u

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

t sin (t)

\frac{d}{d x} (\frac{2}{( 3 x ^{2} + 7 ) ^{2}})

\frac{\partial}{\partial x} (e^{x y} sin (z))

\frac{d}{d x} (4 x^{2 x})

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

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