Integrate the function frac{(log x)^{2}}{x}. | vedclass

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

Question

(A) Let $\log x = t$. Then,differentiating with respect to $x$,we get $\frac{1}{x} dx = dt$. Substituting these into the integral,we have: $\int \frac

Vedclass · Accepted Answer

$\frac{(\log x)^{3}}{3} + C$

Vedclass · Answer

(A) Let $\log x = t$. Then,differentiating with respect to $x$,we get $\frac{1}{x} dx = dt$. Substituting these into the integral,we have: $\int \frac{(\log x)^{2}}{x} dx = \int t^{2} dt$. Using the power rule for integration $\int x^{n} dx = \frac{x^{n+1}}{n+1} + C$,we get: $\int t^{2} dt = \frac{t^{3}}{3} + C$. Substituting $t = \log x$ back into the expression,we obtain: $\frac{(\log x)^{3}}{3} + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{(\log x)^{2}}{x}$.

Similar Questions

If $\int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^{k}+c$,then the value of $k$ is

$\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} \, dx$ is equal to

If $\int \frac{x^3}{\sqrt{1+x^2}} d x=A(1+x^2)^{\frac{3}{2}}+B(1+x^2)^{\frac{1}{2}}+C$,then $A+B=$

$\int \tan x \sec^2 x \sqrt{1 - \tan^2 x} \; dx = $

Let $f(x) = \frac{x}{(1 + x^7)^{1/7}}$ and $g(x) = (f \circ f \circ f \circ f \circ f \circ f \circ f)(x)$. Then $\int x^5 g(x) dx$ equals (where $C$ is the constant of integration):

Vedclass Test Series

Exam Paper Generator

Online Exam Module