What is the integral of ((3x^2)/2+4)^2 ?

The integral of ((3x^2)/2+4)^2 is (9x^5)/(20)+4x^3+16x+C

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integral of ((3x^2)/2+4)^2

\int (\frac{3 x ^{2}}{2} + 4)^{2} d x

\frac{9 x ^{5}}{20} + 4 x^{3} + 16 x + C

Solution steps

\int (\frac{3 x ^{2}}{2} + 4)^{2} d x

Simplify

(\frac{3 x ^{2}}{2} + 4)^{2} : \frac{9 x ^{4}}{4} + 12 x^{2} + 16

= \int \frac{9 x ^{4}}{4} + 12 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 \frac{9 x ^{4}}{4} d x + \int 12 x^{2} d x + \int 16 d x

\int \frac{9 x ^{4}}{4} d x = \frac{9 x ^{5}}{20}

\int 12 x^{2} d x = 4 x^{3}

\int 16 d x = 16 x

= \frac{9 x ^{5}}{20} + 4 x^{3} + 16 x

Add a constant to the solution

= \frac{9 x ^{5}}{20} + 4 x^{3} + 16 x + C

What is the integral of ((3x^2)/2+4)^2 ?
The integral of ((3x^2)/2+4)^2 is (9x^5)/(20)+4x^3+16x+C