What is the integral of (5x^2+4)^2 ?

The integral of (5x^2+4)^2 is 5x^5+(40x^3)/3+16x+C

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integral of (5x^2+4)^2

\int (5 x^{2} + 4)^{2} d x

5 x^{5} + \frac{40 x ^{3}}{3} + 16 x + C

Solution steps

\int (5 x^{2} + 4)^{2} d x

Expand

(5 x^{2} + 4)^{2} : 25 x^{4} + 40 x^{2} + 16

= \int 25 x^{4} + 40 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 25 x^{4} d x + \int 40 x^{2} d x + \int 16 d x

\int 25 x^{4} d x = 5 x^{5}

\int 40 x^{2} d x = \frac{40 x ^{3}}{3}

\int 16 d x = 16 x

= 5 x^{5} + \frac{40 x ^{3}}{3} + 16 x

Add a constant to the solution

= 5 x^{5} + \frac{40 x ^{3}}{3} + 16 x + C

What is the integral of (5x^2+4)^2 ?
The integral of (5x^2+4)^2 is 5x^5+(40x^3)/3+16x+C