Solve ∫ (2x^6+1/3x^5+5/2x^3-8x) wrt x=

$$\int ( 2 x ^ { 6 } + \frac { 1 } { 3 } x ^ { 5 } + \frac { 5 } { 2 } x ^ { 3 } - 8 x ) d x =$$

Evaluate

$$\frac{2x^{7}}{7}+\frac{x^{6}}{18}+\frac{5x^{4}}{8}-4x^{2}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 2x^{6}\mathrm{d}x+\int \frac{x^{5}}{3}\mathrm{d}x+\int \frac{5x^{3}}{2}\mathrm{d}x+\int -8x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{6}\mathrm{d}x+\frac{\int x^{5}\mathrm{d}x}{3}+\frac{5\int x^{3}\mathrm{d}x}{2}-8\int x\mathrm{d}x$$

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 2 times \frac{x^{7}}{7}.

$$\frac{2x^{7}}{7}+\frac{\int x^{5}\mathrm{d}x}{3}+\frac{5\int x^{3}\mathrm{d}x}{2}-8\int x\mathrm{d}x$$

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 \frac{1}{3} times \frac{x^{6}}{6}.

$$\frac{2x^{7}}{7}+\frac{x^{6}}{18}+\frac{5\int x^{3}\mathrm{d}x}{2}-8\int x\mathrm{d}x$$

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 \frac{5}{2} times \frac{x^{4}}{4}.

$$\frac{2x^{7}}{7}+\frac{x^{6}}{18}+\frac{5x^{4}}{8}-8\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -8 times \frac{x^{2}}{2}.

$$\frac{2x^{7}}{7}+\frac{x^{6}}{18}+\frac{5x^{4}}{8}-4x^{2}$$

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.

$$\frac{2x^{7}}{7}+\frac{x^{6}}{18}+\frac{5x^{4}}{8}-4x^{2}+С$$

Differentiate w.r.t. x

$$2x^{6}+\frac{x^{5}}{3}+\frac{5x^{3}}{2}-8x$$