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

The integral of (x^3+1)^2 is (x^7)/7+(x^4)/2+x+C

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integral of (x^3+1)^2

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

\frac{x ^{7}}{7} + \frac{x ^{4}}{2} + x + C

Solution steps

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

Expand

(x^{3} + 1)^{2} : x^{6} + 2 x^{3} + 1

= \int x^{6} + 2 x^{3} + 1 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^{6} d x + \int 2 x^{3} d x + \int 1 d x

\int x^{6} d x = \frac{x ^{7}}{7}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 1 d x = x

= \frac{x ^{7}}{7} + \frac{x ^{4}}{2} + x

Add a constant to the solution

= \frac{x ^{7}}{7} + \frac{x ^{4}}{2} + x + C

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

\int_{2 y}^{2} \int_{0}^{1} e^{x^{2}} d y d x

\frac{d}{d x} (e^{3 x} sin (2 x))

y d x + x^{2} d y = 0

\frac{\partial}{\partial y} (x y^{2} + 5 y - 6 x^{3} + 7)

What is the integral of (x^3+1)^2 ?
The integral of (x^3+1)^2 is (x^7)/7+(x^4)/2+x+C