∫ d+1/alpha(alpha-1)^3 wrt x-এর সমাধান করুন

$$\int \frac { d + 1 } { \alpha ( \alpha - 1 ) ^ { 3 } } d x$$

মূল্যায়ন করুন

$$\frac{dx+x}{\alpha \left(\alpha -1\right)^{3}}+С$$

সমাধানের ধাপগুলো

Find the integral of \frac{d+1}{\alpha \left(\alpha -1\right)^{3}} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{d+1}{\alpha \left(\alpha -1\right)^{3}}x$$

সিমপ্লিফাই।

$$\frac{\left(d+1\right)x}{\alpha \left(\alpha -1\right)^{3}}$$

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.

$$\frac{\left(d+1\right)x}{\alpha \left(\alpha -1\right)^{3}}+С$$

w.r.t. x পার্থক্য করুন

$$\frac{d+1}{\alpha \left(\alpha -1\right)^{3}}$$