Solve ∫ (3t^3-t^2)^2dt

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

Evaluate

$$\frac{9t^{7}}{7}-t^{6}+\frac{t^{5}}{5}+С$$

Solution Steps

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

$$\int 9\left(t^{3}\right)^{2}-6t^{3}t^{2}+\left(t^{2}\right)^{2}\mathrm{d}t$$

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

$$\int 9t^{6}-6t^{3}t^{2}+\left(t^{2}\right)^{2}\mathrm{d}t$$

To multiply powers of the same base, add their exponents. Add 3 and 2 to get 5.

$$\int 9t^{6}-6t^{5}+\left(t^{2}\right)^{2}\mathrm{d}t$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int 9t^{6}-6t^{5}+t^{4}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 9t^{6}\mathrm{d}t+\int -6t^{5}\mathrm{d}t+\int t^{4}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$9\int t^{6}\mathrm{d}t-6\int t^{5}\mathrm{d}t+\int t^{4}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{6}\mathrm{d}t with \frac{t^{7}}{7}. Multiply 9 times \frac{t^{7}}{7}.

$$\frac{9t^{7}}{7}-6\int t^{5}\mathrm{d}t+\int t^{4}\mathrm{d}t$$

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 -6 times \frac{t^{6}}{6}.

$$\frac{9t^{7}}{7}-t^{6}+\int t^{4}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{4}\mathrm{d}t with \frac{t^{5}}{5}.

$$\frac{9t^{7}}{7}-t^{6}+\frac{t^{5}}{5}$$

Simplify.

$$\frac{t^{5}}{5}-t^{6}+\frac{9t^{7}}{7}$$

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.

$$\frac{t^{5}}{5}-t^{6}+\frac{9t^{7}}{7}+С$$

Differentiate w.r.t. t

$$\left(3t-1\right)^{2}t^{4}$$