int frac{dx}{x(x^2+1)^3} = ? | vedclass

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

Question

(A) Let $I = \int \frac{dx}{x(x^2+1)^3}$. Multiply numerator and denominator by $x$: $I = \int \frac{x dx}{x^2(x^2+1)^3}$. Let $x^2 = t$,then $2x dx =

Vedclass · Accepted Answer

$\frac{1}{2(x^2+1)} + \frac{1}{4(x^2+1)^2} + \log \sqrt{\frac{x^2}{x^2+1}} + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x(x^2+1)^3}$. Multiply numerator and denominator by $x$: $I = \int \frac{x dx}{x^2(x^2+1)^3}$. Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{dt}{2}$. $I = \frac{1}{2} \int \frac{dt}{t(t+1)^3}$. Using partial fractions: $\frac{1}{t(t+1)^3} = \frac{A}{t} + \frac{B}{t+1} + \frac{C}{(t+1)^2} + \frac{D}{(t+1)^3}$. Solving for constants: $A=1, B=-1, C=-1, D=-1$. $I = \frac{1}{2} [\int \frac{1}{t} dt - \int \frac{1}{t+1} dt - \int \frac{1}{(t+1)^2} dt - \int \frac{1}{(t+1)^3} dt]$. $I = \frac{1}{2} [\log|t| - \log|t+1| + \frac{1}{t+1} + \frac{1}{2(t+1)^2}] + c$. Substituting $t = x^2$: $I = \frac{1}{2} \log \frac{x^2}{x^2+1} + \frac{1}{2(x^2+1)} + \frac{1}{4(x^2+1)^2} + c$. $I = \log \sqrt{\frac{x^2}{x^2+1}} + \frac{1}{2(x^2+1)} + \frac{1}{4(x^2+1)^2} + c$.

$\int \frac{dx}{x(x^2+1)^3} = ?$

Similar Questions

If $\int \frac{3x+1}{(x-1)^3(x+1)} dx = A \cdot \log \left|\frac{x+1}{x-1}\right| + \frac{B}{x-1} + \frac{C}{(x-1)^2} + D$,then $A+B+C=$

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

$\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x$ is

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

The value of $\int \frac{1}{3 \sin x-\cos x+3} d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module