Let f(x) = int frac{x}{(x^2+1)(x^2+3)} dx. If | vedclass

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

Question

(A) Given $f(x) = \int \frac{x}{(x^2+1)(x^2+3)} dx$. Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{1}{2} dt$. Substituting these into the integral,

Vedclass · Accepted Answer

$\frac{1}{4} \log \left(\frac{1}{3}\right)$

Vedclass · Answer

(A) Given $f(x) = \int \frac{x}{(x^2+1)(x^2+3)} dx$. Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{1}{2} dt$. Substituting these into the integral,we get: $f(x) = \frac{1}{2} \int \frac{1}{(t+1)(t+3)} dt$. Using partial fractions: $\frac{1}{(t+1)(t+3)} = \frac{1}{2} \left( \frac{1}{t+1} - \frac{1}{t+3} \right)$. Thus,$f(x) = \frac{1}{2} \cdot \frac{1}{2} \int \left( \frac{1}{t+1} - \frac{1}{t+3} \right) dt = \frac{1}{4} [\log|t+1| - \log|t+3|] + C = \frac{1}{4} \log \left( \frac{x^2+1}{x^2+3} \right) + C$. Given $f(3) = \frac{1}{4} \log \left( \frac{5}{6} \right)$,we have: $\frac{1}{4} \log \left( \frac{3^2+1}{3^2+3} \right) + C = \frac{1}{4} \log \left( \frac{10}{12} \right) + C = \frac{1}{4} \log \left( \frac{5}{6} \right) + C$. Comparing this with the given value,we find $C = 0$. Therefore,$f(x) = \frac{1}{4} \log \left( \frac{x^2+1}{x^2+3} \right)$. Finally,$f(0) = \frac{1}{4} \log \left( \frac{0^2+1}{0^2+3} \right) = \frac{1}{4} \log \left( \frac{1}{3} \right)$.

Let $f(x) = \int \frac{x}{(x^2+1)(x^2+3)} dx$. If $f(3) = \frac{1}{4} \log \left(\frac{5}{6}\right)$,then find $f(0)$.

Similar Questions

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

If $\int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} - 4)}}} dx = \log {\left( {\frac{{x - 2}}{{x + 2}}} \right)^a}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^b} + c$,then the values of $a$ and $b$ respectively are

$\int \frac{x+1}{x(1+x e^x)^2} d x=$

Integrate the function: $\frac{1}{(x^{2}+1)(x^{2}+4)}$

$\int \frac{\tan x}{\cos x(\sec x-1)(\sec x-2)} d x=$ . . . . . . $+c$

Vedclass Test Series

Exam Paper Generator

Online Exam Module