∫ (x+1)^2 wrt x-এর সমাধান করুন

$$\int ( x + 1 ) ^ { 2 } d x$$

মূল্যায়ন করুন

$$\frac{\left(x+1\right)^{3}}{3}+С$$

সমাধানের ধাপগুলো

\left(x+1\right)^{2} প্রসারিত করতে বাইনোমিয়াল উপপাদ্য \left(a+b\right)^{2}=a^{2}+2ab+b^{2} ব্যবহার করুন৷

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}. 2 কে \frac{x^{2}}{2} বার গুণ করুন।

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

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{x^{3}}{3}+x^{2}+x+С$$

w.r.t. x পার্থক্য করুন

$$\left(x+1\right)^{2}$$