∫ x^2left(x^3+2x^2-5right) wrt x-এর সমাধান করুন

$$\int{ { x }^{ 2 } \left( { x }^{ 3 } +2 { x }^{ 2 } -5 \right) }d x$$

মূল্যায়ন করুন

$$\frac{x^{6}}{6}+\frac{2x^{5}}{5}-\frac{5x^{3}}{3}+С$$

সমাধানের ধাপগুলো

x^{2} কে x^{3}+2x^{2}-5 দিয়ে গুণ করতে ডিস্ট্রিবিউটিভ প্রোপার্টি ব্যবহার করুন।

$$\int x^{5}+2x^{4}-5x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{5}\mathrm{d}x+\int 2x^{4}\mathrm{d}x+\int -5x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{5}\mathrm{d}x+2\int x^{4}\mathrm{d}x-5\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.

$$\frac{x^{6}}{6}+2\int x^{4}\mathrm{d}x-5\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. 2 কে \frac{x^{5}}{5} বার গুণ করুন।

$$\frac{x^{6}}{6}+\frac{2x^{5}}{5}-5\int x^{2}\mathrm{d}x$$

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}. -5 কে \frac{x^{3}}{3} বার গুণ করুন।

$$\frac{x^{6}}{6}+\frac{2x^{5}}{5}-\frac{5x^{3}}{3}$$

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.

$$\frac{x^{6}}{6}+\frac{2x^{5}}{5}-\frac{5x^{3}}{3}+С$$

w.r.t. x পার্থক্য করুন

$$x^{2}\left(x^{3}+2x^{2}-5\right)$$