∫ x^3+x^2-2x wrt x-এর সমাধান করুন

$$\int{ { x }^{ 3 } + { x }^{ 2 } -2x }d x$$

মূল্যায়ন করুন

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}-x^{2}+С$$

সমাধানের ধাপগুলো

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x+\int -2x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}-2\int x\mathrm{d}x$$

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}. -2 কে \frac{x^{2}}{2} বার গুণ করুন।

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}-x^{2}$$

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.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}-x^{2}+С$$

w.r.t. x পার্থক্য করুন

$$x\left(x-1\right)\left(x+2\right)$$