Solve ∫ x^2(x^3+2x^2-5) wrt x

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

Evaluate

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

Solution Steps

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

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 2 times \frac{x^{5}}{5}.

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

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}. Multiply -5 times \frac{x^{3}}{3}.

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

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.

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

Differentiate w.r.t. x

$$x^{2}\left(x^{3}+2x^{2}-5\right)$$