Solve ∫ (x^2+4x)^2 wrt x

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

Evaluate

$$\frac{x^{5}}{5}+2x^{4}+\frac{16x^{3}}{3}+С$$

Solution Steps

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

$$\int \left(x^{2}\right)^{2}+8x^{2}x+16x^{2}\mathrm{d}x$$

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

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

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$\frac{x^{5}}{5}+8\int x^{3}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 8 times \frac{x^{4}}{4}.

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

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}.

$$\frac{x^{5}}{5}+2x^{4}+\frac{16x^{3}}{3}$$

Simplify.

$$\frac{16x^{3}}{3}+2x^{4}+\frac{x^{5}}{5}$$

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.

$$\frac{16x^{3}}{3}+2x^{4}+\frac{x^{5}}{5}+С$$

Differentiate w.r.t. x

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