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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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

\frac{d}{d x} (6 e^{x} (cos (x) - sin (x)))

\int 8 t^{3} d t

\frac{d}{d x} (\frac{1}{x})

x \to 1 + lim (\frac{x ^{2}}{∣ 1 - x ∣})

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