∫ (x+sqrt[3]{x}+1/x^2) wrt x-এর সমাধান করুন

$$\int ( x + \sqrt[ 3 ] { x } + \frac { 1 } { x ^ { 2 } } ) d x$$

মূল্যায়ন করুন

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}+С$$

সমাধানের ধাপগুলো

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

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

$$\frac{x^{2}}{2}+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

x^{\frac{1}{3}} হিসেবে \sqrt[3]{x} পুনরায় লিখুন৷ Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{3}}\mathrm{d}x with \frac{x^{\frac{4}{3}}}{\frac{4}{3}}. সিমপ্লিফাই।

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}+\int \frac{1}{x^{2}}\mathrm{d}x$$

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

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}$$

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.

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}+С$$

w.r.t. x পার্থক্য করুন

$$x+\sqrt[3]{x}+\frac{1}{x^{2}}$$