Solve ∫ x^3(x^4+2x^2+1) wrt x

$$\int x ^ { 3 } ( x ^ { 4 } + 2 x ^ { 2 } + 1 ) d x$$

Evaluate

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

Solution Steps

Use the distributive property to multiply x^{3} by x^{4}+2x^{2}+1.

$$\int x^{7}+2x^{5}+x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{7}\mathrm{d}x+\int 2x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{7}\mathrm{d}x+2\int x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{7}\mathrm{d}x with \frac{x^{8}}{8}.

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

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 2 times \frac{x^{6}}{6}.

$$\frac{x^{8}}{8}+\frac{x^{6}}{3}+\int x^{3}\mathrm{d}x$$

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}.

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

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.

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

Differentiate w.r.t. x

$$x^{3}\left(x^{2}+1\right)^{2}$$