Solve ∫ (x^4+4/sqrt{x}+4) wrt x

$$\int ( x ^ { 4 } + \frac { 4 } { \sqrt { x } } + 4 ) d x$$

Evaluate

$$\frac{x^{5}}{5}+4x+8\sqrt{x}+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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. Multiply 4 times 2\sqrt{x}.

$$\frac{x^{5}}{5}+8\sqrt{x}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}+8\sqrt{x}+4x$$

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.

$$\frac{x^{5}}{5}+8\sqrt{x}+4x+С$$

Differentiate w.r.t. x

$$x^{4}+4+\frac{4}{\sqrt{x}}$$