∫ 1/x^2 wrt x-এর সমাধান করুন

$$\int{ \frac{ 1 }{ { x }^{ 2 } } }d x$$

মূল্যায়ন করুন

$$-\frac{1}{x}+С$$

সমাধানের ধাপগুলো

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}.

$$-\frac{1}{x}$$

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.

$$-\frac{1}{x}+С$$

w.r.t. x পার্থক্য করুন

$$\frac{1}{x^{2}}$$