∫ 3x^2+x wrt x=-এর সমাধান করুন

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

মূল্যায়ন করুন

$$x^{3}+\frac{x^{2}}{2}+С$$

সমাধানের ধাপগুলো

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}. 3 কে \frac{x^{3}}{3} বার গুণ করুন।

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

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}.

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

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.

$$x^{3}+\frac{x^{2}}{2}+С$$

w.r.t. x পার্থক্য করুন

$$x\left(3x+1\right)$$