If x({x^4} + 1)phi (x) = 1, then int_1^2 {phi | vedclass

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

Question

(A) Given $x({x^4} + 1)\phi (x) = 1,$ we have $\phi (x) = \frac{1}{{x({x^4} + 1)}}.$ We can write $\phi (x)$ using partial fractions as: $\phi (x) = \

Vedclass · Accepted Answer

$\frac{1}{4}\log \frac{{32}}{{17}}$

Vedclass · Answer

(A) Given $x({x^4} + 1)\phi (x) = 1,$ we have $\phi (x) = \frac{1}{{x({x^4} + 1)}}.$ We can write $\phi (x)$ using partial fractions as: $\phi (x) = \frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}.$ Now,we integrate $\phi (x)$ from $1$ to $2$: $\int_1^2 {\phi (x)\,dx = \int_1^2 {\left( {\frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}} \right)\,dx} }.$ $= \left[ \log |x| \right]_1^2 - \int_1^2 {\frac{{{x^3}}}{{{x^4} + 1}}\,dx}.$ Let $u = {x^4} + 1,$ then $du = 4{x^3}\,dx,$ so ${x^3}\,dx = \frac{1}{4}du.$ When $x = 1, u = 2.$ When $x = 2, u = 17.$ $= (\log 2 - \log 1) - \frac{1}{4} \int_2^{17} {\frac{1}{u}\,du} = \log 2 - \frac{1}{4} \left[ \log u \right]_2^{17}.$ $= \log 2 - \frac{1}{4} (\log 17 - \log 2) = \log 2 - \frac{1}{4} \log 17 + \frac{1}{4} \log 2.$ $= \frac{5}{4} \log 2 - \frac{1}{4} \log 17 = \frac{1}{4} (5 \log 2 - \log 17) = \frac{1}{4} \log \left( \frac{2^5}{17} \right) = \frac{1}{4} \log \frac{32}{17}.$ Thus,the correct option is $A$.

If $x({x^4} + 1)\phi (x) = 1,$ then $\int_1^2 {\phi (x)\,dx = } $

Similar Questions

Suppose $g(x) = \int_0^x f(t) dt$,where $f$ is such that for $t \in [0, 1]$,$0 \le f(t) \le \frac{1}{2}$ and for $t \in [1, 2]$,$\frac{1}{2} \le f(t) \le 1$. Find the range of $g(2)$.

$\int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2}+4 \sin \frac{x}{2}} \, dx$ is equal to

$\int_{-2}^3 |1-x^2| dx =$

If $f(t) = \int_{-t}^{t} \frac{dx}{1 + x^2}$,then $f'(1)$ is

$\int_0^3 \frac{3x+1}{x^2+9} dx$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module