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

Which ion binds with troponin during muscle contraction?

$A$ radioactive material decays by simultaneous emission of two particles with respective half-lives $1620 \, years$ and $810 \, years$. The time, in years, after which one-fourth of the material remains is

If $f(x)$ satisfies the relation $f\left( \frac{5x - 3y}{2} \right) = \frac{5f(x) - 3f(y)}{2}$ for all $x, y \in R$ and $f(0)=1, f'(0)=2$,then the period of $\sin(f(x))$ is:

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^9+\beta^9$ is equal to

What is preserved in a museum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module