int frac{1}{xleft[6(log x)^2+7 log x+2right]} | vedclass

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

Question

(B) $I = \int \frac{1}{x[6(\log x)^2 + 7 \log x + 2]} dx$Let $t = \log x$,then $dt = \frac{1}{x} dx$.Substituting these into the integral,we get:$I =

Vedclass · Accepted Answer

$\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C$

Vedclass · Answer

(B) $I = \int \frac{1}{x[6(\log x)^2 + 7 \log x + 2]} dx$Let $t = \log x$,then $dt = \frac{1}{x} dx$.Substituting these into the integral,we get:$I = \int \frac{dt}{6t^2 + 7t + 2}$Factorizing the denominator: $6t^2 + 7t + 2 = 6t^2 + 4t + 3t + 2 = 2t(3t + 2) + 1(3t + 2) = (2t + 1)(3t + 2)$.Using partial fractions: $\frac{1}{(3t + 2)(2t + 1)} = \frac{A}{3t + 2} + \frac{B}{2t + 1}$$1 = A(2t + 1) + B(3t + 2)$For $t = -\frac{1}{2}$,$1 = B(3(-\frac{1}{2}) + 2) = B(-\frac{3}{2} + 2) = B(\frac{1}{2}) \implies B = 2$.For $t = -\frac{2}{3}$,$1 = A(2(-\frac{2}{3}) + 1) = A(-\frac{4}{3} + 1) = A(-\frac{1}{3}) \implies A = -3$.Thus,$I = \int (\frac{-3}{3t + 2} + \frac{2}{2t + 1}) dt$$I = -\frac{3 \log |3t + 2|}{3} + \frac{2 \log |2t + 1|}{2} + C$$I = -\log |3t + 2| + \log |2t + 1| + C$$I = \log |\frac{2t + 1}{3t + 2}| + C$Substituting $t = \log x$ back,we get:$I = \log |\frac{2 \log x + 1}{3 \log x + 2}| + C$

$\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x$ is

Similar Questions

$\int \frac{dx}{x(x^2+1)^3} = ?$

$\frac{x^4}{(x^2+1)(x^2+3)} =$

If $\int \frac{3x+1}{(x-1)(x-2)(x-3)} dx = A \log |x-1| + B \log |x-2| + C \log |x-3| + K$,then the values of $A, B$,and $C$ are respectively:

$\int {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}} \,dx = $

If $\int \frac{2 x^2-3}{\left(x^2-4\right)\left(x^2+1\right)} d x=A \tan^{-1} x+B \log (x-2)+C \log (x+2)$,then $6 A+7 B-5 C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module