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Differentiate w.r.t. x
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\int \sqrt[7]{x}\mathrm{d}x+\int -\frac{1}{x^{3}}\mathrm{d}x+\int -\frac{1}{x^{4}}\mathrm{d}x+\int \frac{1}{\sqrt{x}}\mathrm{d}x
Integrate the sum term by term.
\int \sqrt[7]{x}\mathrm{d}x-\int \frac{1}{x^{3}}\mathrm{d}x-\int \frac{1}{x^{4}}\mathrm{d}x+\int \frac{1}{\sqrt{x}}\mathrm{d}x
Factor out the constant in each of the terms.
\frac{7x^{\frac{8}{7}}}{8}-\int \frac{1}{x^{3}}\mathrm{d}x-\int \frac{1}{x^{4}}\mathrm{d}x+\int \frac{1}{\sqrt{x}}\mathrm{d}x
Rewrite \sqrt[7]{x} as x^{\frac{1}{7}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{7}}\mathrm{d}x with \frac{x^{\frac{8}{7}}}{\frac{8}{7}}. Simplify.
\frac{7x^{\frac{8}{7}}}{8}+\frac{1}{2x^{2}}-\int \frac{1}{x^{4}}\mathrm{d}x+\int \frac{1}{\sqrt{x}}\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^{3}}\mathrm{d}x with -\frac{1}{2x^{2}}. Multiply -1 times -\frac{1}{2x^{2}}.
\frac{7x^{\frac{8}{7}}}{8}+\frac{1}{2x^{2}}+\frac{1}{3x^{3}}+\int \frac{1}{\sqrt{x}}\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^{4}}\mathrm{d}x with -\frac{1}{3x^{3}}. Multiply -1 times -\frac{1}{3x^{3}}.
\frac{7x^{\frac{8}{7}}}{8}+\frac{1}{2x^{2}}+\frac{1}{3x^{3}}+2\sqrt{x}
Rewrite \frac{1}{\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{1}{2}}}{\frac{1}{2}}. Simplify and convert from exponential to radical form.
\frac{7x^{\frac{8}{7}}}{8}+2\sqrt{x}+\frac{\frac{1}{3}+\frac{x}{2}}{x^{3}}
Simplify.
\frac{7x^{\frac{8}{7}}}{8}+2\sqrt{x}+\frac{\frac{1}{3}+\frac{x}{2}}{x^{3}}+С
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.