∫ (12x^3-6x^2+8x-3) wrt x=-এর সমাধান করুন

$$\int ( 12 x ^ { 3 } - 6 x ^ { 2 } + 8 x - 3 ) d x =$$

মূল্যায়ন করুন

$$3x^{4}-2x^{3}+4x^{2}-3x+С$$

সমাধানের ধাপগুলো

Integrate the sum term by term.

$$\int 12x^{3}\mathrm{d}x+\int -6x^{2}\mathrm{d}x+\int 8x\mathrm{d}x+\int -3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$12\int x^{3}\mathrm{d}x-6\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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}. 12 কে \frac{x^{4}}{4} বার গুণ করুন।

$$3x^{4}-6\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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}. -6 কে \frac{x^{3}}{3} বার গুণ করুন।

$$3x^{4}-2x^{3}+8\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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}. 8 কে \frac{x^{2}}{2} বার গুণ করুন।

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

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

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

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.

$$3x^{4}-2x^{3}+4x^{2}-3x+С$$

w.r.t. x পার্থক্য করুন

$$12x^{3}-6x^{2}+8x-3$$