حل ∫ x^5+x^3-20x/x+2 wrt x

$$\int{ \frac{ { x }^{ 5 } + { x }^{ 3 } -20x }{ x+2 } }d x$$

تقدير القيمة

$$\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{5x^{3}}{3}-5x^{2}+С$$

خطوات الحل

حدد عوامل التعبيرات التي لم يتم تحديد عواملها بالفعل في \frac{x^{5}+x^{3}-20x}{x+2}.

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

حذف x+2 في البسط والمقام.

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

توسيع التعبير.

$$\int x^{4}-2x^{3}+5x^{2}-10x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -2x^{3}\mathrm{d}x+\int 5x^{2}\mathrm{d}x+\int -10x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-2\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-10\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-2\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-10\int x\mathrm{d}x$$

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}. اضرب -2 في \frac{x^{4}}{4}.

$$\frac{x^{5}}{5}-\frac{x^{4}}{2}+5\int x^{2}\mathrm{d}x-10\int x\mathrm{d}x$$

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}. اضرب 5 في \frac{x^{3}}{3}.

$$\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{5x^{3}}{3}-10\int x\mathrm{d}x$$

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}. اضرب -10 في \frac{x^{2}}{2}.

$$\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{5x^{3}}{3}-5x^{2}$$

تبسيط.

$$-5x^{2}+\frac{5x^{3}}{3}-\frac{x^{4}}{2}+\frac{x^{5}}{5}$$

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.

$$-5x^{2}+\frac{5x^{3}}{3}-\frac{x^{4}}{2}+\frac{x^{5}}{5}+С$$

تفاضل w.r.t. x

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