Solve ∫ 5x^3-1+x^4

$$\int 5 x ^ { 3 } - 1 + x ^ { 4 }$$

Evaluate

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

Solution Steps

Integrate the sum term by term.

$$\int 5x^{3}\mathrm{d}x+\int -1\mathrm{d}x+\int x^{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$5\int x^{3}\mathrm{d}x+\int -1\mathrm{d}x+\int x^{4}\mathrm{d}x$$

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}.

$$\frac{5x^{4}}{4}+\int -1\mathrm{d}x+\int x^{4}\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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}.

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

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.

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

Differentiate w.r.t. x

$$x^{4}+5x^{3}-1$$