Solve ∫ (4z^2-9)^2 wrt z

$$\int ( 4 z ^ { 2 } - 9 ) ^ { 2 } d z$$

Evaluate

$$\frac{16z^{5}}{5}-24z^{3}+81z+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4z^{2}-9\right)^{2}.

$$\int 16\left(z^{2}\right)^{2}-72z^{2}+81\mathrm{d}z$$

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

$$\int 16z^{4}-72z^{2}+81\mathrm{d}z$$

Integrate the sum term by term.

$$\int 16z^{4}\mathrm{d}z+\int -72z^{2}\mathrm{d}z+\int 81\mathrm{d}z$$

Factor out the constant in each of the terms.

$$16\int z^{4}\mathrm{d}z-72\int z^{2}\mathrm{d}z+\int 81\mathrm{d}z$$

Since \int z^{k}\mathrm{d}z=\frac{z^{k+1}}{k+1} for k\neq -1, replace \int z^{4}\mathrm{d}z with \frac{z^{5}}{5}. Multiply 16 times \frac{z^{5}}{5}.

$$\frac{16z^{5}}{5}-72\int z^{2}\mathrm{d}z+\int 81\mathrm{d}z$$

Since \int z^{k}\mathrm{d}z=\frac{z^{k+1}}{k+1} for k\neq -1, replace \int z^{2}\mathrm{d}z with \frac{z^{3}}{3}. Multiply -72 times \frac{z^{3}}{3}.

$$\frac{16z^{5}}{5}-24z^{3}+\int 81\mathrm{d}z$$

Find the integral of 81 using the table of common integrals rule \int a\mathrm{d}z=az.

$$\frac{16z^{5}}{5}-24z^{3}+81z$$

Simplify.

$$81z-24z^{3}+\frac{16z^{5}}{5}$$

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

$$81z-24z^{3}+\frac{16z^{5}}{5}+С$$

Differentiate w.r.t. z

$$\left(4z^{2}-9\right)^{2}$$