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

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

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

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

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

Solution steps

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

Expand

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

g (x) = \frac{5}{6 x ^{3}}

x^{2 - 3 x} x^{2}

\int_{1}^{3} \frac{x ^{2} + 6}{4 x - x ^{2}} d x

\int \frac{4 + t + t ^{2}}{t} d t

y = \frac{1}{x - 2}, at x = 6

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