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

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

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

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

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

Solution steps

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

Expand

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

= \int x^{5} + x^{3} 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^{5} d x + \int x^{3} d x

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

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

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

Add a constant to the solution

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

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

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

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

\frac{d y}{d x} = sin (3 x) - cos (5 x)

(3 t - 1) (4 t - 4)^{- 1}

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