∫ (1/x-sinx+3) wrt x-এর সমাধান করুন

$$\int ( \frac { 1 } { x } - \sin x + 3 ) d x$$

মূল্যায়ন করুন

$$\ln(|x|)+\cos(x)+3x+С$$

সমাধানের ধাপগুলো

Integrate the sum term by term.

$$\int \frac{1}{x}\mathrm{d}x+\int -\sin(x)\mathrm{d}x+\int 3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int \frac{1}{x}\mathrm{d}x-\int \sin(x)\mathrm{d}x+\int 3\mathrm{d}x$$

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

$$\ln(|x|)-\int \sin(x)\mathrm{d}x+\int 3\mathrm{d}x$$

Use \int \sin(x)\mathrm{d}x=-\cos(x) from the table of common integrals to obtain the result. -1 কে -\cos(x) বার গুণ করুন।

$$\ln(|x|)+\cos(x)+\int 3\mathrm{d}x$$

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

$$\ln(|x|)+\cos(x)+3x$$

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.

$$\ln(|x|)+\cos(x)+3x+С$$

w.r.t. x পার্থক্য করুন

$$-\sin(x)+3+\frac{1}{x}$$