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

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

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

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

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

Solution steps

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

Expand

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

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

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 t^{4} d t - \int 4 t^{2} d t + \int 1 d t

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

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

\int 1 d t = t

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

Add a constant to the solution

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

2 (x) + 1

- 7 x^{2} + 63, - x^{2} + 9

x \to - 1 lim (\frac{1 - x ^{2}}{x + 1})

\int (4 x^{3} - 1)^{2} d x

f (x) = x^{\frac{1}{5}}

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