If int frac{log (1+x^4)}{x^3} d x=f(x) log le | vedclass

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

Question

(B) We are given the integral $I = \int \frac{\log (1+x^4)}{x^3} d x$. Using integration by parts,let $u = \log (1+x^4)$ and $dv = x^{-3} dx$. Then $d

Vedclass · Accepted Answer

$\frac{g(x)}{2}$

Vedclass · Answer

(B) We are given the integral $I = \int \frac{\log (1+x^4)}{x^3} d x$. Using integration by parts,let $u = \log (1+x^4)$ and $dv = x^{-3} dx$. Then $du = \frac{4x^3}{1+x^4} dx$ and $v = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$. $I = -\frac{1}{2x^2} \log (1+x^4) - \int \left(-\frac{1}{2x^2}ight) \frac{4x^3}{1+x^4} dx = -\frac{1}{2x^2} \log (1+x^4) + \int \frac{2x}{1+x^4} dx$. Let $t = x^2$,then $dt = 2x dx$. $I = -\frac{1}{2x^2} \log (1+x^4) + \int \frac{dt}{1+t^2} = -\frac{1}{2x^2} \log (1+x^4) + 	an^{-1}(x^2) + c$. Comparing this with $f(x) \log \left(\frac{1}{g(x)}ight) + 	an^{-1}(h(x)) + c$,we have $f(x) = \frac{1}{2x^2}$,$g(x) = 1+x^4$,and $h(x) = x^2$. Now,$f\left(\frac{1}{x}ight) = \frac{1}{2(1/x)^2} = \frac{x^2}{2}$. Then $h(x) \left[f(x) + f\left(\frac{1}{x}ight)ight] = x^2 \left(\frac{1}{2x^2} + \frac{x^2}{2}ight) = \frac{1}{2} + \frac{x^4}{2} = \frac{1+x^4}{2} = \frac{g(x)}{2}$.

If $\int \frac{\log (1+x^4)}{x^3} d x=f(x) \log \left(\frac{1}{g(x)}\right)+\tan ^{-1}(h(x))+c$,then $h(x)\left[f(x)+f\left(\frac{1}{x}\right)\right]=$

Similar Questions

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

Find the following integral: $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$

If $\int \frac{1 + \sqrt{\tan x}}{\sin 2x} dx = A \log \tan x + B \sqrt{\tan x} + C$ then $4A - B =$

If $\int e^u \sin 2x \, dx$ can be expressed in terms of elementary functions of $x$,then $u$ can be:

$\int \frac{2 x+2}{\sqrt{x^2-4 x-5}} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module