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

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

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integral of (1+y^3)^2

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d y = y

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

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

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

Add a constant to the solution

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

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