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

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(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,$I = \frac{1}{2} \int \left(\frac{2}{t+2} - \frac{1}{t+1}\right) dt$. $I = \frac{1}{2} [2 \log|t+2| - \log|t+1|] + c$. $I = \log|x^2+2| - \frac{1}{2} \log|x^2+1| + c = \log \left(\frac{x^2+2}{\sqrt{x^2+1}}\right) + c$.

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

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