What is the integral of (2x^2+4)^2 ?

The integral of (2x^2+4)^2 is (4x^5)/5+(16x^3)/3+16x+C

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integral of (2x^2+4)^2

\int (2 x^{2} + 4)^{2} d x

\frac{4 x ^{5}}{5} + \frac{16 x ^{3}}{3} + 16 x + C

Solution steps

\int (2 x^{2} + 4)^{2} d x

Expand

(2 x^{2} + 4)^{2} : 4 x^{4} + 16 x^{2} + 16

= \int 4 x^{4} + 16 x^{2} + 16 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 4 x^{4} d x + \int 16 x^{2} d x + \int 16 d x

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

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

\int 16 d x = 16 x

= \frac{4 x ^{5}}{5} + \frac{16 x ^{3}}{3} + 16 x

Add a constant to the solution

= \frac{4 x ^{5}}{5} + \frac{16 x ^{3}}{3} + 16 x + C

What is the integral of (2x^2+4)^2 ?
The integral of (2x^2+4)^2 is (4x^5)/5+(16x^3)/3+16x+C