If int frac{d x}{x(log x-2)(log x-3)}=I+C,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I_1 = \int \frac{d x}{x(\log x-2)(\log x-3)}$.Substitute $t = \log x$,then $d t = \frac{1}{x} d x$.The integral becomes $I_1 = \int \frac{d t

Vedclass · Accepted Answer

$\log \left|\frac{\log x-3}{\log x-2}\right|$

Vedclass · Answer

(B) Let $I_1 = \int \frac{d x}{x(\log x-2)(\log x-3)}$.Substitute $t = \log x$,then $d t = \frac{1}{x} d x$.The integral becomes $I_1 = \int \frac{d t}{(t-2)(t-3)}$.Using partial fractions,$\frac{1}{(t-2)(t-3)} = \frac{A}{t-2} + \frac{B}{t-3}$.$1 = A(t-3) + B(t-2)$.For $t=3$,$B=1$. For $t=2$,$A=-1$.So,$I_1 = \int \left( \frac{1}{t-3} - \frac{1}{t-2} \right) d t$.$I_1 = \log |t-3| - \log |t-2| + C = \log \left| \frac{t-3}{t-2} \right| + C$.Substituting $t = \log x$ back,we get $I = \log \left| \frac{\log x - 3}{\log x - 2} \right|$.

If $\int \frac{d x}{x(\log x-2)(\log x-3)}=I+C$,then $I$ is equal to

Similar Questions

In the circuit shown in the figure,$R = \sqrt{\frac{L}{C}}$. Switch $S$ is closed at time $t = 0$. The current through the inductor $L$ and the capacitor $C$ would be equal after a time $t$ equal to:

$HCl$ is added to the following oxides,which one would give $H_2O_2$?

The value of $\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} - 1 = $

Which of the following statements is correct?

Assertion : The isothermal curves intersect each other at a certain point.
Reason : The isothermal change takes place slowly,so,the isothermal curves have very little slope.

Vedclass Test Series

Exam Paper Generator

Online Exam Module