Solve ∫ (2t+5x)(2t-5x)

$$\int ( 2 t + 5 x ) ( 2 t - 5 x )$$

Evaluate

$$4xt^{2}-\frac{25x^{3}}{3}+С$$

Solution Steps

Consider \left(2t+5x\right)\left(2t-5x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(2t\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\int 4t^{2}-\left(5x\right)^{2}\mathrm{d}x$$

Expand \left(5x\right)^{2}.

$$\int 4t^{2}-5^{2}x^{2}\mathrm{d}x$$

Calculate 5 to the power of 2 and get 25.

$$\int 4t^{2}-25x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4t^{2}\mathrm{d}x+\int -25x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int t^{2}\mathrm{d}x-25\int x^{2}\mathrm{d}x$$

Find the integral of t^{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$4t^{2}x-25\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -25 times \frac{x^{3}}{3}.

$$4t^{2}x-\frac{25x^{3}}{3}$$

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

$$4t^{2}x-\frac{25x^{3}}{3}+С$$

Differentiate w.r.t. x

$$4t^{2}-25x^{2}$$