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

The integral of (x^3+1)^2x^2 is (x^9)/9+(x^6)/3+(x^3)/3+C

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

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

\frac{x ^{9}}{9} + \frac{x ^{6}}{3} + \frac{x ^{3}}{3} + C

Solution steps

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

Expand

(x^{3} + 1)^{2} x^{2} : x^{8} + 2 x^{5} + x^{2}

= \int x^{8} + 2 x^{5} + x^{2} 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^{8} d x + \int 2 x^{5} d x + \int x^{2} d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 2 x^{5} d x = \frac{x ^{6}}{3}

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

= \frac{x ^{9}}{9} + \frac{x ^{6}}{3} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{9}}{9} + \frac{x ^{6}}{3} + \frac{x ^{3}}{3} + C

(x + 1) y^{^{'}} + (x + 2) y = 2 x e^{- x}

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 10, (27, 1)

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

6 x^{4} - 9 x y - 7 y^{2} = 104, (- 2, 2)

n = 1 \sum \infty \frac{- n}{n ^{2} + 1}

What is the integral of (x^3+1)^2x^2 ?
The integral of (x^3+1)^2x^2 is (x^9)/9+(x^6)/3+(x^3)/3+C