Solve = ∫ (7x^6-5x^3+2x^2+3) wrt x

$$= \int ( 7 x ^ { 6 } - 5 x ^ { 3 } + 2 x ^ { 2 } + 3 ) d x$$

Evaluate

$$x^{7}-\frac{5x^{4}}{4}+\frac{2x^{3}}{3}+3x+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}. Multiply -5 times \frac{x^{4}}{4}.

$$x^{7}-\frac{5x^{4}}{4}+2\int x^{2}\mathrm{d}x+\int 3\mathrm{d}x$$

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

$$x^{7}-\frac{5x^{4}}{4}+\frac{2x^{3}}{3}+\int 3\mathrm{d}x$$

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{7}-\frac{5x^{4}}{4}+\frac{2x^{3}}{3}+3x$$

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.

$$x^{7}-\frac{5x^{4}}{4}+\frac{2x^{3}}{3}+3x+С$$

Differentiate w.r.t. x

$$7x^{6}-5x^{3}+2x^{2}+3$$