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

Integrate the rational function: $\frac{\cos x}{(1-\sin x)(2-\sin x)}$
[Hint: Put $\sin x = t$]

Let $I(x) = \int \frac{(x+1)}{x(1+x e^x)^2} dx, x > 0$. If $\lim_{x \rightarrow \infty} I(x) = 0$,then $I(1)$ is equal to

For $x > 1$,evaluate the integral: $\int \frac{1}{x(x^4 - 1)} \, dx$

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

$\int \frac{dx}{x - x^2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module