int frac{x^3}{x^4+5 x^2+4} ,d x= | vedclass

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Question

(A) Let $I = \int \frac{x^3}{x^4+5x^2+4} dx$. Substitute $t = x^2$,then $dt = 2x dx$,or $x dx = \frac{1}{2} dt$. The integral becomes $I = \frac{1}{2}

Vedclass · Accepted Answer

$\frac{1}{3} \log \left(\frac{\left(x^2+4\right)^2}{\sqrt{x^2+1}}\right)+c$,where $c$ is the constant of integration

Vedclass · Answer

(A) Let $I = \int \frac{x^3}{x^4+5x^2+4} dx$. Substitute $t = x^2$,then $dt = 2x dx$,or $x dx = \frac{1}{2} dt$. The integral becomes $I = \frac{1}{2} \int \frac{t}{t^2+5t+4} dt$. Factor the denominator: $t^2+5t+4 = (t+1)(t+4)$. Using partial fractions: $\frac{t}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$. $t = A(t+4) + B(t+1)$. For $t = -1$,$-1 = 3A \implies A = -1/3$. For $t = -4$,$-4 = -3B \implies B = 4/3$. So,$I = \frac{1}{2} \int \left( \frac{4/3}{t+4} - \frac{1/3}{t+1} \right) dt = \frac{1}{6} \int \left( \frac{4}{t+4} - \frac{1}{t+1} \right) dt$. $I = \frac{1}{6} [4 \log|t+4| - \log|t+1|] + c$. $I = \frac{1}{6} \log \left| \frac{(t+4)^4}{t+1} \right| + c$. Substituting $t = x^2$ back: $I = \frac{1}{6} \log \left( \frac{(x^2+4)^4}{x^2+1} \right) + c = \frac{1}{3} \log \left( \frac{(x^2+4)^2}{\sqrt{x^2+1}} \right) + c$.

$\int \frac{x^3}{x^4+5 x^2+4} \,d x=$

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