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

The integral of (x^3+1)^2x is (x^8)/8+(2x^5)/5+(x^2)/2+C

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

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

\frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

(x^{3} + 1)^{2} x : x^{7} + 2 x^{4} + x

= \int x^{7} + 2 x^{4} + x 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^{7} d x + \int 2 x^{4} d x + \int x d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

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

= \frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2} + C

What is the integral of (x^3+1)^2x ?
The integral of (x^3+1)^2x is (x^8)/8+(2x^5)/5+(x^2)/2+C