int frac{ln |x|}{xsqrt{1 + ln |x|}} , dx equa | vedclass

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

Question

(A) Let $t = 1 + \ln |x|$. Then $dt = \frac{1}{x} \, dx$,and $\ln |x| = t - 1$.Substituting these into the integral:$\int \frac{t - 1}{\sqrt{t}} \, dt

Vedclass · Accepted Answer

$\frac{2}{3} \sqrt{1 + \ln |x|} (\ln |x| - 2) + c$

Vedclass · Answer

(A) Let $t = 1 + \ln |x|$. Then $dt = \frac{1}{x} \, dx$,and $\ln |x| = t - 1$.Substituting these into the integral:$\int \frac{t - 1}{\sqrt{t}} \, dt = \int (\sqrt{t} - \frac{1}{\sqrt{t}}) \, dt$$= \int (t^{1/2} - t^{-1/2}) \, dt$$= \frac{t^{3/2}}{3/2} - \frac{t^{1/2}}{1/2} + c$$= \frac{2}{3} t^{3/2} - 2 t^{1/2} + c$$= \frac{2}{3} \sqrt{t} (t - 3) + c$Substituting $t = 1 + \ln |x|$ back:$= \frac{2}{3} \sqrt{1 + \ln |x|} (1 + \ln |x| - 3) + c$$= \frac{2}{3} \sqrt{1 + \ln |x|} (\ln |x| - 2) + c$.

$\int \frac{\ln |x|}{x\sqrt{1 + \ln |x|}} \, dx$ equals :

Similar Questions

$\int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} = $

Integrate the function: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$

$\int \frac{dx}{2\sqrt{x}(1 + x)} = $

$\int {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}} \,dx$ equals

To evaluate $\int \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \, dx$,the most suitable substitution is

Vedclass Test Series

Exam Paper Generator

Online Exam Module