Լուծեք = ∫ 5+2x+1/x wrt x

$$= \int 5 + 2 x + \frac { 1 } { x } d x$$

Գնահատել

$$\ln(|x|)+x^{2}+5x+С$$

Լուծման քայլերը

Integrate the sum term by term.

$$\int 5\mathrm{d}x+\int 2x\mathrm{d}x+\int \frac{1}{x}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 5\mathrm{d}x+2\int x\mathrm{d}x+\int \frac{1}{x}\mathrm{d}x$$

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$5x+2\int x\mathrm{d}x+\int \frac{1}{x}\mathrm{d}x$$

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

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

Use \int \frac{1}{x}\mathrm{d}x=\ln(|x|) from the table of common integrals to obtain the result.

$$5x+x^{2}+\ln(|x|)$$

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.

$$5x+x^{2}+\ln(|x|)+С$$

Տարբերակել վերագրած x-ը

$$2x+5+\frac{1}{x}$$