The value of int frac{dx}{xsqrt{x^4 - 1}} is | vedclass

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

Question

(A) Let $I = \int \frac{dx}{x\sqrt{x^4 - 1}}$.Multiply the numerator and denominator by $x$:$I = \int \frac{x \, dx}{x^2 \sqrt{(x^2)^2 - 1}}$.Substitu

Vedclass · Accepted Answer

$\frac{1}{2} \sec^{-1}(x^2) + k$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x\sqrt{x^4 - 1}}$.Multiply the numerator and denominator by $x$:$I = \int \frac{x \, dx}{x^2 \sqrt{(x^2)^2 - 1}}$.Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.Substituting these into the integral:$I = \int \frac{dt/2}{t \sqrt{t^2 - 1}} = \frac{1}{2} \int \frac{dt}{t \sqrt{t^2 - 1}}$.Using the standard integral formula $\int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + k$:$I = \frac{1}{2} \sec^{-1}(t) + k$.Substituting back $t = x^2$:$I = \frac{1}{2} \sec^{-1}(x^2) + k$.

The value of $\int \frac{dx}{x\sqrt{x^4 - 1}}$ is

Similar Questions

$\int \sec x \, dx = $

$\int {\frac{{\cos x - 1}}{{\cos x + 1}}\,dx} = $

$\int \sec^2 x \csc^2 x \, dx = $ . . . . . . $+ C$

If $\int \frac{f(x) \, dx}{\log \sin x} = \log \log \sin x$,then $f(x) = $

If $\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}} d x=x+E x^{5 / 6}+D x^{2 / 3}+C x^{1 / 2}+B x^{1 / 3}+A x^{1 / 6}+\log (\sqrt[6]{x}-1)^6+K$,then $A+B+C+D+E=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module