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

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

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integral of (y^2+1)^2

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

\frac{y ^{5}}{5} + \frac{2 y ^{3}}{3} + y + C

Solution steps

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

Expand

(y^{2} + 1)^{2} : y^{4} + 2 y^{2} + 1

= \int y^{4} + 2 y^{2} + 1 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 y^{4} d y + \int 2 y^{2} d y + \int 1 d y

\int y^{4} d y = \frac{y ^{5}}{5}

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

\int 1 d y = y

= \frac{y ^{5}}{5} + \frac{2 y ^{3}}{3} + y

Add a constant to the solution

= \frac{y ^{5}}{5} + \frac{2 y ^{3}}{3} + y + C

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