Solve ∫ x^2(2x^3+4)^2 wrt x

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

Evaluate

$$\frac{4x^{9}}{9}+\frac{8x^{6}}{3}+\frac{16x^{3}}{3}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x^{3}+4\right)^{2}.

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

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

$$\int x^{2}\left(4x^{6}+16x^{3}+16\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2} by 4x^{6}+16x^{3}+16.

$$\int 4x^{8}+16x^{5}+16x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{8}\mathrm{d}x+\int 16x^{5}\mathrm{d}x+\int 16x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{8}\mathrm{d}x+16\int x^{5}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{8}\mathrm{d}x with \frac{x^{9}}{9}. Multiply 4 times \frac{x^{9}}{9}.

$$\frac{4x^{9}}{9}+16\int x^{5}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply 16 times \frac{x^{6}}{6}.

$$\frac{4x^{9}}{9}+\frac{8x^{6}}{3}+16\int x^{2}\mathrm{d}x$$

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

$$\frac{4x^{9}}{9}+\frac{8x^{6}}{3}+\frac{16x^{3}}{3}$$

Simplify.

$$\frac{16x^{3}}{3}+\frac{8x^{6}}{3}+\frac{4x^{9}}{9}$$

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.

$$\frac{16x^{3}}{3}+\frac{8x^{6}}{3}+\frac{4x^{9}}{9}+С$$

Differentiate w.r.t. x

$$4\left(x\left(x^{3}+2\right)\right)^{2}$$