Solve ∫ 3x^2(2x^2+4)^2 wrt x

$$\int 3 x ^ { 2 } ( 2 x ^ { 2 } + 4 ) ^ { 2 } d x$$

Evaluate

$$\frac{12x^{7}}{7}+\frac{48x^{5}}{5}+16x^{3}+С$$

Solution Steps

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

$$\int 3x^{2}\left(4\left(x^{2}\right)^{2}+16x^{2}+16\right)\mathrm{d}x$$

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

$$\int 3x^{2}\left(4x^{4}+16x^{2}+16\right)\mathrm{d}x$$

Use the distributive property to multiply 3x^{2} by 4x^{4}+16x^{2}+16.

$$\int 12x^{6}+48x^{4}+48x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 12x^{6}\mathrm{d}x+\int 48x^{4}\mathrm{d}x+\int 48x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$12\int x^{6}\mathrm{d}x+48\int x^{4}\mathrm{d}x+48\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 12 times \frac{x^{7}}{7}.

$$\frac{12x^{7}}{7}+48\int x^{4}\mathrm{d}x+48\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 48 times \frac{x^{5}}{5}.

$$\frac{12x^{7}}{7}+\frac{48x^{5}}{5}+48\int x^{2}\mathrm{d}x$$

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 48 times \frac{x^{3}}{3}.

$$\frac{12x^{7}}{7}+\frac{48x^{5}}{5}+16x^{3}$$

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.

$$\frac{12x^{7}}{7}+\frac{48x^{5}}{5}+16x^{3}+С$$

Differentiate w.r.t. x

$$12\left(x\left(x^{2}+2\right)\right)^{2}$$