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

The integral of t^2(t^3-4)^2 is (t^9)/9-(4t^6)/3+(16t^3)/3+C

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

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

\frac{t ^{9}}{9} - \frac{4 t ^{6}}{3} + \frac{16 t ^{3}}{3} + C

Solution steps

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

Expand

t^{2} (t^{3} - 4)^{2} : t^{8} - 8 t^{5} + 16 t^{2}

= \int t^{8} - 8 t^{5} + 16 t^{2} 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 t^{8} d t - \int 8 t^{5} d t + \int 16 t^{2} d t

\int t^{8} d t = \frac{t ^{9}}{9}

\int 8 t^{5} d t = \frac{4 t ^{6}}{3}

\int 16 t^{2} d t = \frac{16 t ^{3}}{3}

= \frac{t ^{9}}{9} - \frac{4 t ^{6}}{3} + \frac{16 t ^{3}}{3}

Add a constant to the solution

= \frac{t ^{9}}{9} - \frac{4 t ^{6}}{3} + \frac{16 t ^{3}}{3} + C

(x - 2)^{4} + e^{x - 2} + (x - 2 \cdot z)^{2}

\frac{d y}{d x} = x + 7 y

\int 27 x^{26} d x

\frac{- 5 s}{( s ^{2} + 1 ) ^{2}}

x \to 1 lim (\frac{3}{x - 1})

What is the integral of t^2(t^3-4)^2 ?
The integral of t^2(t^3-4)^2 is (t^9)/9-(4t^6)/3+(16t^3)/3+C