Solve ∫ (2t^2-1)^2dt=

$$\int ( 2 t ^ { 2 } - 1 ) ^ { 2 } d t =$$

Evaluate

$$\frac{4t^{5}}{5}-\frac{4t^{3}}{3}+t+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2t^{2}-1\right)^{2}.

$$\int 4\left(t^{2}\right)^{2}-4t^{2}+1\mathrm{d}t$$

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

$$\int 4t^{4}-4t^{2}+1\mathrm{d}t$$

Integrate the sum term by term.

$$\int 4t^{4}\mathrm{d}t+\int -4t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

Factor out the constant in each of the terms.

$$4\int t^{4}\mathrm{d}t-4\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

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}. Multiply 4 times \frac{t^{5}}{5}.

$$\frac{4t^{5}}{5}-4\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

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 -4 times \frac{t^{3}}{3}.

$$\frac{4t^{5}}{5}-\frac{4t^{3}}{3}+\int 1\mathrm{d}t$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{4t^{5}}{5}-\frac{4t^{3}}{3}+t$$

Simplify.

$$t-\frac{4t^{3}}{3}+\frac{4t^{5}}{5}$$

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.

$$t-\frac{4t^{3}}{3}+\frac{4t^{5}}{5}+С$$

Differentiate w.r.t. t

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