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

The integral of (5+4x^2)^2 is 25x+(40x^3)/3+(16x^5)/5+C

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

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

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

Solution steps

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

Expand

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

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

\int 25 d x = 25 x

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

\int 16 x^{4} d x = \frac{16 x ^{5}}{5}

= 25 x + \frac{40 x ^{3}}{3} + \frac{16 x ^{5}}{5}

Add a constant to the solution

= 25 x + \frac{40 x ^{3}}{3} + \frac{16 x ^{5}}{5} + C

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What is the integral of (5+4x^2)^2 ?
The integral of (5+4x^2)^2 is 25x+(40x^3)/3+(16x^5)/5+C