Solve ∫ (x^8-1/4x^6+3x^2+100) wrt x

$$\int ( x ^ { 8 } - \frac { 1 } { 4 } x ^ { 6 } + 3 x ^ { 2 } + 100 ) d x$$

Evaluate

$$\frac{x^{9}}{9}-\frac{x^{7}}{28}+x^{3}+100x+С$$

Solution Steps

Integrate the sum term by term.

$$\int x^{8}\mathrm{d}x+\int -\frac{x^{6}}{4}\mathrm{d}x+\int 3x^{2}\mathrm{d}x+\int 100\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{8}\mathrm{d}x-\frac{\int x^{6}\mathrm{d}x}{4}+3\int x^{2}\mathrm{d}x+\int 100\mathrm{d}x$$

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}.

$$\frac{x^{9}}{9}-\frac{\int x^{6}\mathrm{d}x}{4}+3\int x^{2}\mathrm{d}x+\int 100\mathrm{d}x$$

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 -\frac{1}{4} times \frac{x^{7}}{7}.

$$\frac{x^{9}}{9}-\frac{x^{7}}{28}+3\int x^{2}\mathrm{d}x+\int 100\mathrm{d}x$$

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

$$\frac{x^{9}}{9}-\frac{x^{7}}{28}+x^{3}+\int 100\mathrm{d}x$$

Find the integral of 100 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{9}}{9}-\frac{x^{7}}{28}+x^{3}+100x$$

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.

$$\frac{x^{9}}{9}-\frac{x^{7}}{28}+x^{3}+100x+С$$

Differentiate w.r.t. x

$$x^{8}-\frac{x^{6}}{4}+3x^{2}+100$$