int frac{x^3}{x^4+3 x^2+2} d x= | vedclass

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

Question

(A) Let $I = \int \frac{x^3}{x^4+3 x^2+2} d x$. Substitute $t = x^2$, then $dt = 2x dx$, which implies $x dx = \frac{dt}{2}$. The integral becomes $I

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{x^3}{x^4+3 x^2+2} d x$. Substitute $t = x^2$, then $dt = 2x dx$, which implies $x dx = \frac{dt}{2}$. The integral becomes $I = \int \frac{t}{t^2+3t+2} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{t}{(t+1)(t+2)} dt$. Using partial fractions: $\frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}$. Solving for $A$ and $B$: $t = A(t+2) + B(t+1)$. For $t = -1$, $A = -1$. For $t = -2$, $B = 2$. So, $I = \frac{1}{2} \int \left(\frac{2}{t+2} - \frac{1}{t+1}\right) dt$. $I = \frac{1}{2} [2 \log|t+2| - \log|t+1|] + c$. $I = \log|x^2+2| - \frac{1}{2} \log|x^2+1| + c = \log \left(\frac{x^2+2}{\sqrt{x^2+1}}\right) + c$.

$\int \frac{x^3}{x^4+3 x^2+2} d x=$

Similar Questions

Integrate the rational function: $\frac{1}{x^{4}-1}$

The correct evaluation of $\int \frac{x}{(x - 2)(x - 1)} \, dx$ is (where $p$ is an arbitrary constant):

Integrate the rational function: $\frac{1-x^{2}}{x(1-2 x)}$

If $\int \frac{2x^2+3}{(x^2-1)(x^2-4)} dx = \log \left[\left(\frac{x-2}{x+2}\right)^a \cdot \left(\frac{x+1}{x-1}\right)^b\right] + c$,(where $c$ is the constant of integration),then the value of $a+b$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module