Solve ∫ t^2(t^3-4)^2dt

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

Evaluate

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

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t^{3}-4\right)^{2}.

$$\int t^{2}\left(\left(t^{3}\right)^{2}-8t^{3}+16\right)\mathrm{d}t$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\int t^{2}\left(t^{6}-8t^{3}+16\right)\mathrm{d}t$$

Use the distributive property to multiply t^{2} by t^{6}-8t^{3}+16.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{8}\mathrm{d}t with \frac{t^{9}}{9}.

$$\frac{t^{9}}{9}-8\int t^{5}\mathrm{d}t+16\int t^{2}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{5}\mathrm{d}t with \frac{t^{6}}{6}. Multiply -8 times \frac{t^{6}}{6}.

$$\frac{t^{9}}{9}-\frac{4t^{6}}{3}+16\int t^{2}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}. Multiply 16 times \frac{t^{3}}{3}.

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

Simplify.

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

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

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

Differentiate w.r.t. t

$$\left(t\left(t^{3}-4\right)\right)^{2}$$