Solve ∫ (10x^4+1/x^2+6sqrt{x}-3) wrt x

$$\int{ (10 { x }^{ 4 } + \frac{ 1 }{ { x }^{ 2 } } +6 \sqrt{ x } -3) }d x$$

Evaluate

$$2x^{5}+4x^{\frac{3}{2}}-3x-\frac{1}{x}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 10x^{4}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x+\int 6\sqrt{x}\mathrm{d}x+\int -3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$10\int x^{4}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x+6\int \sqrt{x}\mathrm{d}x+\int -3\mathrm{d}x$$

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

$$2x^{5}+\int \frac{1}{x^{2}}\mathrm{d}x+6\int \sqrt{x}\mathrm{d}x+\int -3\mathrm{d}x$$

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

$$2x^{5}-\frac{1}{x}+6\int \sqrt{x}\mathrm{d}x+\int -3\mathrm{d}x$$

Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify. Multiply 6 times \frac{2x^{\frac{3}{2}}}{3}.

$$2x^{5}-\frac{1}{x}+4x^{\frac{3}{2}}+\int -3\mathrm{d}x$$

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{5}-\frac{1}{x}+4x^{\frac{3}{2}}-3x$$

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.

$$2x^{5}-\frac{1}{x}+4x^{\frac{3}{2}}-3x+С$$

Differentiate w.r.t. x

$$10x^{4}+6\sqrt{x}-3+\frac{1}{x^{2}}$$