What is the integral of (4t^2-3)^2 ?

The integral of (4t^2-3)^2 is (16t^5)/5-8t^3+9t+C

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integral of (4t^2-3)^2

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

\frac{16 t ^{5}}{5} - 8 t^{3} + 9 t + C

Solution steps

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

Expand

(4 t^{2} - 3)^{2} : 16 t^{4} - 24 t^{2} + 9

= \int 16 t^{4} - 24 t^{2} + 9 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 16 t^{4} d t - \int 24 t^{2} d t + \int 9 d t

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

\int 24 t^{2} d t = 8 t^{3}

\int 9 d t = 9 t

= \frac{16 t ^{5}}{5} - 8 t^{3} + 9 t

Add a constant to the solution

= \frac{16 t ^{5}}{5} - 8 t^{3} + 9 t + C

f (x) = x^{2}, [1, 5]

\frac{8 x ^{2} + 2}{x ^{2}}

(x^{2} - 4 x + 5) 25 - x^{2}

\int t - 2 cos (t) d t

x \to 1 lim (sin (\frac{1}{x - 1}))

What is the integral of (4t^2-3)^2 ?
The integral of (4t^2-3)^2 is (16t^5)/5-8t^3+9t+C