Solve ∫ 3x^4-2x^3+7 wrt x=

$$\int 3 x ^ { 4 } - 2 x ^ { 3 } + 7 d x =$$

Evaluate

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

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -2 times \frac{x^{4}}{4}.

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

Find the integral of 7 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Differentiate w.r.t. x

$$3x^{4}-2x^{3}+7$$