Q. 71 - Calculus

Question

Prove $\int x^{k}dx=\frac{1}{k+1}{x}^{k+1}+C if k\varnothing 1 and \int \frac{1}{x}d\left(x\right)=\ln \left(x\right)+C$

Step-by-Step Solution

Answer

Hence Proved

Step 1. Given

The proving equation is $\int x^{k}dx=\frac{1}{k+1}{x}^{k+1}+C if k\varnothing 1 and \int \frac{1}{x}d\left(x\right)=\ln \left(x\right)+C$.

Step 2. Finding derivative

$$\begin{gathered} \mathrm{Finding} \mathrm{derivative} \mathrm{of} \frac{1}{k+1}{x}^{k+1} \\ d\left(\frac{1}{k+1}{x}^{k+1}\right) \\ =\frac{1}{k+1}(k+1)x^{k+1-1} \\ =x^k \\ \mathrm{Since} \frac{1}{k+1}{x}^{k+1} \mathrm{is} \mathrm{antiderivative} \mathrm{of} x^k \\ \mathrm{Therefore} \int x^{k}d\left(x\right)=\frac{1}{k+1}{x}^{k+1} \end{gathered}$$

Step 3. Proving

$$\begin{gathered} \int \frac{1}{x}d\left(x\right)=\ln \left(x\right)+C \\ Now, \\ \frac{d}{dx}\left(\ln \right(x\left)\right) \\ =\frac{1}{x} \\ Since , \ln \left(x\right) is an antidervative of \frac{1}{x} \\ Therefore \int \frac{1}{x}=\ln \left(x\right)+C \end{gathered}$$