Solve ∫ x^2(20x^7-7x^4+1) wrt x

$$\int x ^ { 2 } ( 20 x ^ { 7 } - 7 x ^ { 4 } + 1 ) d x$$

Evaluate

$$2x^{10}-x^{7}+\frac{x^{3}}{3}+С$$

Solution Steps

Use the distributive property to multiply x^{2} by 20x^{7}-7x^{4}+1.

$$\int 20x^{9}-7x^{6}+x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 20x^{9}\mathrm{d}x+\int -7x^{6}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$20\int x^{9}\mathrm{d}x-7\int x^{6}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{9}\mathrm{d}x with \frac{x^{10}}{10}. Multiply 20 times \frac{x^{10}}{10}.

$$2x^{10}-7\int x^{6}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply -7 times \frac{x^{7}}{7}.

$$2x^{10}-x^{7}+\int x^{2}\mathrm{d}x$$

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}.

$$2x^{10}-x^{7}+\frac{x^{3}}{3}$$

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.

$$2x^{10}-x^{7}+\frac{x^{3}}{3}+С$$

Differentiate w.r.t. x

$$20x^{9}-7x^{6}+x^{2}$$