What is the integral of (2x^2-1)^2 ?

The integral of (2x^2-1)^2 is (4x^5)/5-(4x^3)/3+x+C

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integral of (2x^2-1)^2

\int (2 x^{2} - 1)^{2} d x

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

Solution steps

\int (2 x^{2} - 1)^{2} d x

Expand

(2 x^{2} - 1)^{2} : 4 x^{4} - 4 x^{2} + 1

= \int 4 x^{4} - 4 x^{2} + 1 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 4 x^{2} d x + \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{- 2 x} cos (π y))

\int \frac{2 x}{1} d x

n = 0 \sum \infty 1.0 2^{n}

(3, - 3), (0, 2)

\int \frac{x ^{3} - 3 x ^{2} + 4 x - 9}{x ^{2} + 3} d x

What is the integral of (2x^2-1)^2 ?
The integral of (2x^2-1)^2 is (4x^5)/5-(4x^3)/3+x+C