int frac{d x}{xleft(x^4+1right)}= | vedclass

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(B) $\int \frac{d x}{x\left(x^4+1\right)} = \int \frac{x^3 d x}{x^4\left(x^4+1\right)}$Let $x^4 = t$, then $4x^3 dx = dt$, so $x^3 dx = \frac{dt}{4}$.

Vedclass · Accepted Answer

$\frac{1}{4} \log \left(\frac{x^4}{x^4+1}\right)+C$

Vedclass · Answer

(B) $\int \frac{d x}{x\left(x^4+1\right)} = \int \frac{x^3 d x}{x^4\left(x^4+1\right)}$Let $x^4 = t$, then $4x^3 dx = dt$, so $x^3 dx = \frac{dt}{4}$.Substituting these into the integral:$\int \frac{dt/4}{t(t+1)} = \frac{1}{4} \int \frac{dt}{t(t+1)}$Using partial fractions:$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$Therefore:$\frac{1}{4} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt = \frac{1}{4} (\log |t| - \log |t+1|) + C$$= \frac{1}{4} \log \left| \frac{t}{t+1} \right| + C$Substituting $t = x^4$ back:$= \frac{1}{4} \log \left( \frac{x^4}{x^4+1} \right) + C$

$\int \frac{d x}{x\left(x^4+1\right)}=$

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