Solve ∫ x^3(1-5x^2) wrt x

$$\int x ^ { 3 } ( 1 - 5 x ^ { 2 } ) d x$$

Evaluate

$$-\frac{5x^{6}}{6}+\frac{x^{4}}{4}+С$$

Solution Steps

Use the distributive property to multiply x^{3} by 1-5x^{2}.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{4}}{4}-5\int x^{5}\mathrm{d}x$$

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 -5 times \frac{x^{6}}{6}.

$$\frac{x^{4}}{4}-\frac{5x^{6}}{6}$$

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.

$$\frac{x^{4}}{4}-\frac{5x^{6}}{6}+С$$

Differentiate w.r.t. x

$$x^{3}-5x^{5}$$