What is the integral of (4t^2-5)^2 ?

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

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integral of (4t^2-5)^2

\int (4 t^{2} - 5)^{2} d t

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

Solution steps

\int (4 t^{2} - 5)^{2} d t

Expand

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

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

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

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

\int 25 d t = 25 t

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

Add a constant to the solution

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

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